4 More about Linear Functions

4.1 Systems of Linear Equations

Consider the following set of data on the world record times R , in seconds, in the 100 meter dash for both men and women at various times t over the last century.

1940

1950

1960

1970

1980

1990

2000

Men

10.2

10.2

10.0

9.95

9.95

9.9

9.84

Women

11.6

11.5

11.3

11.0

10.88

10.49

10.49

Regression analysis gives us the two linear functions that best fit these world record times:

for men:

Equationupper R equals minus 0.0062 t plus 10.438

for women:

Equationupper R equals minus 0.0206 t plus 12.48 comma

where t represents the number of years since 1900. The corresponding scatterplot with the two lines superimposed is shown in Figure 4.1. Both lines are slanted downward, as we would expect, but the line for the women has a considerably larger negative slope. In fact, when we compare the slopes, we see that the slope

Equationleft-parenthesis m equals negative 0.0206 right-parenthesis
for the women's record time is more than three times as large as the slope
Equationleft-parenthesis m equals negative 0.0062 right-parenthesis
for the men. This observation suggests an obvious question: If these trends continue, when will the women's world record surpass the men's world record?

Figure 4.1

This image shows a line graph that compares the world record times in the 100 meter dash for both men and women. The X axis shows the years and the Y Axis shows the time. The data shown is taken from the table above.

To answer this question, we have to find the point where the extensions of the two lines intersect one another. We can estimate this point numerically by trial-and-error (substitute different values for t into both equations until we find the time t when the women's world record is less than the men's record). For instance, we can try

Equationt equals 70
,
Equationt equals 80
, etc. and eventually narrow things down to get a more and more accurate value. Alternatively, we can also estimate the point of intersection graphically by extending the two lines and finding the point where they cross. As seen in Figure 4.2, the two lines seem to cross when t is between 141 and 142, which means roughly 141 years after 1900, or around 2041. We can obviously zoom in on the graph to obtain a more accurate estimate for the point of intersection.

The third approach is to think of the two equations as a pair of simultaneous linear equations . To do so, it is probably a good idea to rewrite both equations by bringing the variable terms to the same side of each equation by adding the appropriate quantities to both sides of each equation. When we do this, we obtain

Equationupper R plus 0.0062 t equals 10.438

Equationupper R plus 0.0206 t equals 12.48 period

Before trying to solve this pair of equations, we first need to consider some fundamental ideas on solving 40 systems of linear equations. We will come back to this situation later in this section.

Figure 4.2

This image is a line graph with the data shown previously. However, the x axis does not stop at 110, but continues until 160. The lines continue until they intersect at 140. No extra data has been plotted.

This image is of a statue used as a bullet point

Thought Experiment

Use trial-and-error to estimate the year, correct to one decimal place, in which the women's world record time in the 100 meter dash will surpass the men's record, assuming that the trends in the world records continue.

A pair of linear equations such as

Equationx plus 5 y equals negative 3 (1)

Equation4 x minus 3 y equals 11 (2)

is called a system of simultaneous linear equations or simply a system of linear equations . Its solution is a pair of values, one for x and the other for y , that satisfy both equations simultaneously . The solution to the system of Equations (1) and (2) is Equationx equals 2 and Equationy equals negative 1 (you will see below how we get it). To verify that this is indeed the solution, we substitute Equationx equals 2 and Equationy equals negative 1 into each of the two original equations in turn. For Equation (1), we have

Equationleft-parenthesis 2 right-parenthesis plus 5 left-parenthesis minus 1 right-parenthesis equals 2 minus 5 equals negative 3

and for Equation (2),

Equation4 left-parenthesis 2 right-parenthesis minus 3 left-parenthesis negative 1 right-parenthesis equals 8 plus 3 equals 11
,

so these two values satisfy both equations and hence is the solution.

Solving Systems of Linear Equations Geometrically

For now, we consider only systems of two linear equations in two unknowns. Geometrically, each of the two linear equations represents a line and every pair of numbers x and y that satisfies either of the two equations is a point (x , y ) on that line. A single pair of values for x and y that satisfy both equations simultaneously must be the point of intersection of the two lines, as shown in Figure 4.3. To graph these lines using a calculator or a computer program, it is typically necessary to solve for y in terms of x . For Equation (1), we first subtract x from both sides to get

Equation5 y equals negative x minus 3

and then divide both sides by 5 to obtain

Equationy equals StartFraction negative x minus 3 Over 5 EndFraction equals minus one-fifth x minus three-fifths period

For Equation (2), we first add 3y to both sides to get

Equation4 x equals 3 y plus 11

and then subtract 11 from both sides, so that

Equation4 x minus 11 equals 3 y period

When we divide both sides by 3, we have

Equationy equals StartFraction 4 x minus 11 Over 3 EndFraction equals four-thirds x minus StartFraction 11 Over 3 EndFraction period

From Figure 4.3, we see that the two lines seem to intersect at the point

Equationleft-parenthesis 2 comma negative 1 right-parenthesis
, which is the solution we cited above.

Figure 4.3

This image shows two linear equations plotted on a graph. The x axis ranges from Minus 3 to 5 and the y axis ranges from Minus 4 to 3. One line has an equation of x plus 5 y Equals Minus 3. The other line has an equation of 4 x Minus 3 y Equals 11. The two lines intersect at left paren 2 comma Minus 1 right paren.

This image is of a statue used as a bullet point

Thought Experiment

Use the graphical approach to estimate the year in which the women's world record time in the 100 meter dash will surpass the men's record assuming that the trends in the world records continue.

We can always use this graphical approach to solve systems of two equations in two unknowns, but the best we can get is a reasonably accurate estimate of the solution, even after zooming in on the intersection point repeatedly. Worse, if we have a system of more than two equations in two unknowns, say a system of four equations in four unknowns, then such a geometric approach is impossible. Alternatively, we can solve a system of linear equations algebraically.

Solving Systems of Linear Equations Algebraically

We begin by reviewing briefly some ideas on solving a system of linear equations algebraically, as a reminder and a demonstration of pencil-and-paper techniques. Later in this section, we will look at the far more efficient and easier methods that are used in practice today.

Typically, the algebraic approach involves one of two methods, the method of substitution and the method of elimination .

1. The method of substitution: Solve for one variable (say x ) in terms of the other variable (in this case, y ) using either of the two equations. Next, substitute the expression for that variable (x ) into the other equation, leaving a single equation in the other variable (y ). (This is usually straightforward if the coefficient of one of the variables is 1 or Equationnegative 1; otherwise it can get fairly messy.) Then solve for the first variable. Finally, substitute its value back into either of the previous equations to find the value of the other variable. We illustrate this process in the following example.

Example 1 Solve the system of Equations (1) and (2) using the method of substitution.

Solution Since the coefficient of x in Equation (1) is 1, we use Equation (1) to solve for x in terms of y as

Equationx equals minus 5 y minus 3. (3)

We substitute this expression into Equation (2) to get

Equation4 left-parenthesis minus 5 y minus 3 right-parenthesis minus 3 y equals 11
.

We apply the distributive law to obtain

Equationminus 20 y minus 12 minus 3 y equals 11
Equationa left-parenthesis u plus v right-parenthesis equals a u plus a v

Note that we have eliminated the variable x and now have a single equation in y only. We collect like terms to get

Equationminus 23 y minus 12 equals 11

Equationminus 23 y
Equationequals 23
,

so that, when we divide both sides by

Equationnegative 23
,

Equationy equals 23 slash left-parenthesis negative 23 right-parenthesis equals negative 1
.

We now substitute

Equationy equals negative 1
back into Equation (3) to find that

Equationx equals minus 5 left-parenthesis negative 1 right-parenthesis minus 3

Equationequals 5 negative 3 equals 2
.

This is the same solution

Equationx equals 2
and
Equationy equals negative 1
we obtained graphically in Figure 4.3. (Note that we could have substituted
Equationy equals negative 1
into Equation (1) or Equation (2) instead of Equation (3), but that would have resulted in more complicated expressions.)

2. The method of elimination: Add or subtract an appropriate multiple of one of the equations to the other equation to eliminate one of the variables.

Example 2 Solve the same system of linear equations

Equationx plus 5 y equals negative 3
(1)

Equation4 x minus 3 y equals 11
(2)

using the method of elimination.

Solution Suppose we choose to eliminate the variable x from the two equations. To do this, we multiply Equation (1) by

Equationnegative 4
to get

Equationminus 4 x minus 20 y equals 12
(4)

while Equation (2) is

Equation4 x minus 3 y equals 11 period
(2)

(We did this to produce coefficients of x that are numerically equal, but of opposite sign.) If we add Equation (2) and Equation (4), the x terms cancel, leaving

Equationminus 23 y equals 23 comma

so that, when we divide both sides by Equationnegative 23,

Equationy equals negative 1,

as before. To solve for the other variable x , we now substitute this value for y into either Equation (1) or (2), say Equation (1), and get

Equationx plus 5 left-parenthesis negative 1 right-parenthesis equals negative 3

or

Equationx minus 5 equals negative 3

so that, when we add 5 to both sides, we get

Equationx equals 5 minus 3 equals 2,

the same solution once more. Note that we would have obtained the same value for x if we had substituted

Equationy equals negative 1
into Equation (2) instead.

Alternatively, instead of eliminating x from Equations (1) and (2), we could eliminate the variable y . To do this, we multiply Equation (1) by 3 and Equation (2) by 5 to get:

Equation3 times upper E q n left-parenthesis 1 right-parenthesis
:
Equation3 x plus 15 y equals negative 9

Equation5 times upper E q n left-parenthesis 2 right-parenthesis
:
Equation20 x minus 15 y equals 55 period

Notice that one equation has the term

Equation15 y
and the other has the term
Equationminus 15 y
. We eliminate y by adding these two equations to get

Equation23 x equals 46
,

and so

Equationx equals 2
. Substituting
Equationx equals 2
into either Equation (1) or Equation (2) gives
Equationy equals negative 1
.

Example 3 Solve the system of linear equations

Equationupper R plus 0.0062 t equals 10.438

Equationupper R plus 0.0206 t equals 12.48

for the world record times in the 100 meter dash for men and for women using (a) the method of substitution; (b) the method of elimination.

Solution a. From the two equations, it is obviously simpler to solve for R from either equation and substitute into the other equation than to solve for t . Using the first equation, we have

Equationupper R equals 10.438 minus 0.0062 t
,

so when we substitute this into the second equation, we get

Equationupper R plus 0.0206 t equals 10.438 minus 0.0062 t plus 0.0206 t equals 12.48

so that

Equation10.438 plus 0.0144 t equals 12.48
.

Combining like terms, we then have

Equation0.0144 t equals 12.48 minus 10.438 equals 2.042
,

so that

Equationt equals 2.042 slash 0.0144 equals 141.80556 almost-equals 141.8 years

or almost 142 years after 1900, which would be in late 2042. This value agrees with our graphical solution at the beginning of the section.

b. To solve the system of equations using the elimination method, we subtract the first equation from the second equation (to have a positive coefficient for the t -term) and so obtain

Equation0.0206 t minus 0.0062 t equals 12.48 minus 10.438
,

which becomes

Equation0.0144 t equals 2.042

and therefore

Equationt equals 2.042 slash 0.0144 equals 141.80555 almost-equals 141.8 years
,

which is the identical solution.

If we have a system of three linear equations in three unknowns (say x , y , and z or

Equationx 1
,
Equationx 2
, and
Equationx 3
or u , v , and w ), let alone an even larger system of linear equations (such as four equations in four unknowns), the method of substitution quickly becomes far too complicated. The method of elimination can be used but, in practice, once we have systems larger than two-by-two, it is standard to use technology to find the solution rather than to attempt to solve the system by hand. We illustrate the algebraic approach in Appendix I for anyone who would like to see how it is done.

Solving Systems of Linear Equations Using Matrices

Most graphing calculators have the capability of solving systems of up to 99 equations in 99 unknowns at the push of a button. On some calculators, there is a SIMULT key for simultaneous equations; you typically enter the number of linear equations, and then enter the coefficients and the constant terms, and finally press Solve to get the solutions. On other calculators, you can solve a system of linear equations using the Solve command; you enter the list of equations, the list of variables, and press enter. On all scientific or more sophisticated calculators, you can solve systems of equations using matrix methods, as discussed below. The routines used by the calculators and by software packages to solve systems of linear equations all use matrix methods.

A matrix is any rectangular array of numbers, such as

EquationStart 3 By 4 Matrix 1st Row 1st Column 4 2nd Column 0 3rd Column 5 4th Column 1 2nd Row 1st Column 7 2nd Column negative 2 3rd Column 8 4th Column 6 3rd Row 1st Column negative 3 2nd Column 1 3rd Column 2 4th Column 3 EndMatrix period

The size, or dimension , of a matrix is measured by the number of rows (horizontally across) and the number of columns (vertically down) in the array. Since the matrix above has 3 rows and 4 columns, its dimension is 3 by 4, which we can write as

Equation3 times 4
. In this book, we will only consider two kinds of matrices:

square matrices , say 2 by 2 or 3 by 3 or 4 by 4, and

column matrices that have only a single column, say 2 rows and 1 column or 4 rows and 1 column. Such column matrices are called column vectors or simply vectors .

Most of the ideas and methods we discuss extend in a very natural way to square matrices of size n by n , for any positive integer n and vectors of size n by 1, and many of these ideas can also extend to more general rectangular matrices.

Let's look at the system of linear equations:

Equation2 x plus 5 y plus 4 z equals negative 7

Equation3 x minus y plus 5 z equals 14

Equation6 x plus 2 y minus 2 z equals 4

(We consider this same system in Appendix I, where we find the solution algebraically to be

Equationx equals 2
,
Equationy equals negative 3
, and
Equationz equals 1
.) We first construct the coefficient matrix A, which is made up of the coefficients from each of the equations:

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 2 2nd Column 5 3rd Column 4 2nd Row 1st Column 3 2nd Column negative 1 3rd Column 5 3rd Row 1st Column 6 2nd Column 2 3rd Column negative 2 EndMatrix period

This matrix has 3 rows and 3 columns, so its dimension is 3 by 3, or

Equation3 times 3
. We also construct the matrix B of constants, made up of the constants on the right-hand side of the three equations:

Equationupper B equals Start 3 By 1 Matrix 1st Row negative 7 2nd Row 14 3rd Row 4 EndMatrix period

Since this vector has 3 rows and 1 column, its dimension is

Equation3 times 1
. Finally, we construct the
Equation3 times 1
matrix X of variables:

Equationupper X equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix period

These are the three unknowns we must determine.

The system of three linear equations in three unknowns is then equivalent to the simple matrix equation

Equationupper A upper X equals upper B period

For now, we only consider the mechanics of solving this matrix equation; later, we will look at some of the mathematical details. The solution X to the matrix equation

Equationupper A upper X equals upper B
is found in terms of the inverse matrix , which we write as
Equationupper A Superscript negative 1
(assuming it exists), for the matrix A:

Equationupper X equals upper A Superscript negative 1 Baseline upper B period

The solution to the matrix equation

Equationupper A upper X equals upper B

is the matrix

Equationupper X equals upper A Superscript negative 1 Baseline upper B

assuming

Equationupper A Superscript negative 1
exists.

To solve this matrix equation on the calculator, and so solve the corresponding system of linear equations, you must "name" each of the matrices in turn by giving its dimension and then entering the values for each position in each matrix. The dimension of matrix A is Equation3 times 3 and the dimension of matrix B is Equation3 times 1. Finally, by selecting the appropriate names of the matrices, you have the calculator find

Equationupper A Superscript negative 1 Baseline asterisk upper B
and it displays the entries in the column matrix X:

EquationStart 3 By 1 Matrix 1st Row 2 2nd Row negative 3 3rd Row 1 EndMatrix period

See the instruction manual for your calculator for more specific details on how to use its matrix features. We will discuss the meaning of matrix multiplication and the inverse matrix in more detail in Section 4.3; for now, we will focus on the details of solving systems of linear equations and some of the many applications that lead to such systems.

Incidentally, if any of the terms in one or more of the equations is missing, as in

Equation4 x plus 5 z equals 8
, then the corresponding entry in the matrix would be 0 to represent
Equation0 y
. Also, if any of the coefficients have negative signs, as in the coefficients of y and z in
Equation3 x minus 4 y minus 2 z equals 11
, the corresponding entries in the matrix would be
Equationnegative 4
and
Equationnegative 2
.

This image is of a statue used as a bullet point

Thought Experiment

Use the matrix features of your calculator to verify that

Equationx equals 2
,
Equationy equals negative 3
and Equationz equals 1 is the solution of the system

Equation2 x plus 5 y plus 4 z equals negative 7

Equation3 x minus y plus 5 z equals 14

Equation6 x plus 2 y minus 2 z equals 4

E xample 4 Use matrices to solve the system of linear equations

Equation5 x minus 7 y plus 4 z equals 6

Equation2 x plus 4 y minus 8 z equals 13

Equation3 x minus 5 y minus 9 z equals negative 2

Solution The coefficient matrix and the matrix of constants are

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 5 2nd Column negative 7 3rd Column 4 2nd Row 1st Column 2 2nd Column 4 3rd Column negative 8 3rd Row 1st Column 3 2nd Column negative 5 3rd Column negative 9 EndMatrix and Equationupper B equals Start 3 By 1 Matrix 1st Row 6 2nd Row 13 3rd Row negative 2 EndMatrix.

When we enter these matrices into the calculator and form the expression Equationupper A Superscript negative 1 Baseline asterisk upper B (or simply Equationupper A Superscript negative 1 Baseline upper B), we find that the corresponding matrix of variables is

Equationupper X equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix equals upper A Superscript negative 1 Baseline upper B equals Start 3 By 1 Matrix 1st Row 3.791079812 2nd Row 2.049295775 3rd Row 0.347178404 EndMatrix period

Correct to three decimal places, the solution to the system of equations is Equationx almost-equals 3.791, Equationy almost-equals 2.049, and Equationz almost-equals 0.347 period

E xample 5 Use matrix methods to solve the system of linear equations

Equationupper R plus 0.0062 t equals 10.438

Equationupper R plus 0.0206 t equals 12.48

for the world record times in the 100 meter dash for men and for women.

Solution The coefficient matrix and the matrix of constants are

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0.0062 2nd Row 1st Column 1 2nd Column 0.0206 EndMatrix
and
Equationupper B equals StartBinomialOrMatrix 10.438 Choose 12.48 EndBinomialOrMatrix
.

We enter these matrices into the calculator and form the expression Equationupper A Superscript negative 1 Baseline upper B to find that the corresponding matrix of variables is

Equationupper X equals StartBinomialOrMatrix x Choose y EndBinomialOrMatrix equals StartBinomialOrMatrix upper R Choose t EndBinomialOrMatrix equals upper A Superscript negative 1 Baseline upper B equals StartBinomialOrMatrix 9.55881 Choose 141.80886 EndBinomialOrMatrix period

Thus, when Equationt almost-equals 141.8, the world record time for both men and women would be about

Equationupper R equals 9.559
seconds, assuming that the two linear trends continue that long.

The value of using matrix methods is that they are just as easy to apply to much larger systems of linear equations, as we illustrate in the following example.

E xample 6 Use matrices to solve the following system of linear equations:

Equation2 x minus 5 y plus 3 z minus 2 w equals 10

Equation4 x plus 8 y minus 2 z plus 3 w equals negative 3

Equationminus 3 x minus 4 y minus 3 z plus 4 w equals 2

Equation5 x minus 4 y minus 3 z plus 4 w equals negative 5

Solution For this system of four equations in four unknowns, the coefficient matrix and the matrix of constants are

Equationupper A equals Start 4 By 4 Matrix 1st Row 1st Column 2 2nd Column negative 5 3rd Column 3 4th Column negative 2 2nd Row 1st Column 4 2nd Column 8 3rd Column negative 2 4th Column 3 3rd Row 1st Column negative 3 2nd Column negative 4 3rd Column negative 3 4th Column 4 4th Row 1st Column 5 2nd Column negative 4 3rd Column negative 3 4th Column 4 EndMatrix
and
Equationupper B equals Start 4 By 1 Matrix 1st Row 10 2nd Row negative 3 3rd Row 2 4th Row negative 5 EndMatrix
.

When we enter these matrices into the calculator and form the expression Equationupper A Superscript negative 1 Baseline upper B, we find that the corresponding matrix of variables is

Equationupper X equals Start 4 By 1 Matrix 1st Row x 2nd Row y 3rd Row z 4th Row w EndMatrix equals upper A Superscript negative 1 Baseline upper B equals Start 4 By 1 Matrix 1st Row negative 0.875 2nd Row negative 0.7705 3rd Row 7.26541 4th Row 5.21575 EndMatrix period

That is, correct to three decimal places, the solution to the system of equations is

Equationx almost-equals negative 0.875
,
Equationy almost-equals negative 0.771
,
Equationz almost-equals 7.265
, and
Equationw almost-equals 5.216
.

Problems

1. Verify whether or not

Equationx equals 1
,
Equationy equals 2
, and
Equationz equals 3
is the solution of the system of equations

Equation4 x plus 3 y minus 2 z equals 5

Equation5 x minus 2 y plus 3 z equals 10

Equation3 x minus y minus 2 z equals negative 5

2. Verify whether or not

Equationx equals 1
,
Equationy equals 2
, and
Equationz equals 3
is the solution of the system of equations

Equation4 x plus 3 y minus 2 z equals 4

Equation5 x minus 2 y plus 3 z equals 10

Equation3 x minus y minus 2 z equals 5

3. Verify whether or not

Equationx equals 1
,
Equationy equals 2
, and
Equationz equals 3
is the solution of the system of equations

Equation4 x plus 3 y minus 2 z equals 4

Equation5 x minus 2 y plus 3 z equals 10

Equation3 x minus y minus 2 z equals negative 5

4. Verify whether or not

Equationx equals 2 y equals negative 1
,
Equationz equals 3
, and
Equationw equals 2
is the solution of the system of equations

Equation2 x plus 3 y minus 2 z plus 5 w equals 5

Equation4 x minus 5 y plus 3 z minus 2 w equals 18

Equation6 x minus 2 y minus 4 z plus w equals 4

Equation3 x plus 4 z minus 5 w equals 8

5. What are the dimensions of each of the following matrices?

a. EquationStart 3 By 2 Matrix 1st Row 1st Column 2 2nd Column 3 2nd Row 1st Column 1 2nd Column 0 3rd Row 1st Column negative 5 2nd Column 4 EndMatrix

b. EquationStart 2 By 4 Matrix 1st Row 1st Column 1 2nd Column 3 3rd Column negative 4 4th Column 6 2nd Row 1st Column 7 2nd Column 0 3rd Column 2 4th Column negative 5 EndMatrix

c. EquationStart 4 By 3 Matrix 1st Row 1st Column 6 2nd Column 1 3rd Column 3 2nd Row 1st Column 5 2nd Column 2 3rd Column negative 4 3rd Row 1st Column 5 2nd Column 0 3rd Column 1 4th Row 1st Column 2 2nd Column negative 7 3rd Column 4 EndMatrix

Problems 6–8 refer to the matrix Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 4 2nd Column 5 3rd Column 1 2nd Row 1st Column 2 2nd Column 2 3rd Column 1 3rd Row 1st Column 0 2nd Column negative 3 3rd Column 2 EndMatrix period In each, two vectors are shown. One is the solution vector X and the other is the vector of constants B in the matrix equation

Equationupper A upper X equals upper B period
Which is which?

6. EquationStart 3 By 1 Matrix 1st Row negative 2 2nd Row 1 3rd Row 2 EndMatrix and EquationStart 3 By 1 Matrix 1st Row negative 1 2nd Row 0 3rd Row 1 EndMatrix

7. EquationStart 3 By 1 Matrix 1st Row 31 2nd Row 14 3rd Row negative 11 EndMatrix and EquationStart 3 By 1 Matrix 1st Row 1 2nd Row 5 3rd Row 2 EndMatrix

8. EquationStart 3 By 1 Matrix 1st Row 23 2nd Row 9 3rd Row negative 12 EndMatrix and EquationStart 3 By 1 Matrix 1st Row 4 2nd Row 2 3rd Row negative 3 EndMatrix

9. Determine, using your calculator or a software package, whether or not each of the following matrices have inverses.

a. EquationStart 2 By 2 Matrix 1st Row 1st Column 2 2nd Column negative 3 2nd Row 1st Column negative 8 2nd Column 12 EndMatrix

b. EquationStart 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 7 2nd Row 1st Column negative 2 2nd Column 5 EndMatrix

c. EquationStart 3 By 3 Matrix 1st Row 1st Column negative 2 2nd Column 0 3rd Column 5 2nd Row 1st Column 1 2nd Column 3 3rd Column 2 3rd Row 1st Column 0 2nd Column 6 3rd Column 9 EndMatrix

d. EquationStart 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 2 3rd Column 3 2nd Row 1st Column 4 2nd Column 5 3rd Column 6 3rd Row 1st Column 7 2nd Column 8 3rd Column 9 EndMatrix

e. EquationStart 4 By 4 Matrix 1st Row 1st Column 2 2nd Column 3 3rd Column negative 2 4th Column 5 2nd Row 1st Column 0 2nd Column 5 3rd Column 7 4th Column 2 3rd Row 1st Column 1 2nd Column negative 3 3rd Column 1 4th Column 4 4th Row 1st Column 6 2nd Column 2 3rd Column 0 4th Column negative 3 EndMatrix

f. EquationStart 4 By 4 Matrix 1st Row 1st Column 2 2nd Column 3 3rd Column negative 2 4th Column 5 2nd Row 1st Column 0 2nd Column 5 3rd Column 7 4th Column 2 3rd Row 1st Column 1 2nd Column negative 3 3rd Column 1 4th Column 4 4th Row 1st Column 3 2nd Column 5 3rd Column 6 4th Column 11 EndMatrix

Find the solution to each of the following systems of equations graphically.

10. Equationx plus y equals 8

Equation3 x minus 3 y equals 4

11. Equationx minus 4 y equals 3

Equationnegative x plus 2 y equals 9

12.

Equation2 x minus 3 y equals 4

Equationminus 4 x plus y equals 9

13.

Equation3 x minus 5 y equals 7

Equation2 x plus 6 y equals 11

14.

Equation5 x plus 4 y equals 8

Equation6 x minus 2 y equals 15

1 5 .

Equation3 x plus 7 y equals 12

Equation6 x plus 3 y equals negative 11

Find the solution, if it exists, to each of the following systems of equations using appropriate technology.

16. Equationx plus y equals 8

Equation2 x minus 3 y equals 4

17. Equationx minus 4 y equals 3

Equationnegative x plus 2 y equals 9

18.

Equation2 x minus 3 y equals 4

Equationminus 4 x plus y equals 1

19. Equation3 a minus 2 b equals 3

Equationminus 2 a plus 5 b equals 2

20. Equation3 u minus 2 v equals 3

Equationminus 4 u plus 3 v equals negative 2

21. Equation20 x plus 4 y plus 4 z equals 500

Equation8 x plus 3 y plus 5 z equals 800

Equation4 x plus 5 y plus 9 z equals 2000

22. Equation6 x plus 5 y plus 6 z equals 500

Equation10 x plus 10 y equals 500

Equation2 x plus 12 z equals 100

23. Equation8 x plus 4 y plus 3 z equals 500

Equation4 x plus 8 y plus 5 z equals 500

Equation12 y plus 6 z equals 500

24. Equation6 x plus 2 y plus 2 z equals 500

Equation3 x plus 6 y plus 3 z equals 500

Equation3 x plus 2 y plus 6 z equals 1000

25. Equationa plus 2 b plus c equals 3

Equationnegative a plus 2 b plus 3 c equals 1

Equation2 a minus b plus 2 c equals 2

26. Equation2 a minus b plus c equals 2

Equationnegative a minus 2 b plus 2 c equals negative 3

Equation3 a plus 2 b minus c equals 0

27. Equationnegative p minus 3 q plus 2 r equals negative 1

Equation5 p plus 4 q plus 6 r equals 12

Equation2 p plus q plus 3 r equals 4

28. Equation2 p plus 4 q minus 2 r equals 4

Equationminus 2 p plus 2 q minus 3 r equals negative 4

Equationp minus q minus 2 r equals negative 1

29. Equationu plus 2 v plus 2 w equals 3

Equationu plus 2 v plus 3 w equals 11

Equation2 u plus v plus 2 w equals 5

30. Equationu minus v minus 2 w equals 2

Equation3 u plus 2 v plus 4 w equals 2

Equation3 u plus v minus 2 w equals negative 3

4.2 Applications of Linear Equations

Systems of linear equations arise in many different ways. We begin with a rather routine application that is typical of the kind of problems you likely saw in high school algebra. However, our primary focus here will be on other kinds of applications and we will consider a much wider range of applications of these ideas later in this section.

E xample 1 One of the standard dinner offerings at Brookdale College's dining hall is meatloaf with mashed potatoes and green beans. The staff dietitian has to make up a meal with 475 calories, 36 grams of protein, and 55 grams of carbs. (These values are called the demand .) Each ounce of meatloaf has 75 calories, 7 grams of protein, and 6 grams of carbs. Each scoop of mashed potatoes has 60 calories, 2 grams of protein, and 10 grams of carbs. Each ounce of green beans has 10 calories, 1 gram of protein, and 2 grams of carbs. The dietician has to determine how many ounces of meatloaf, how many scoops of mashed potatoes, and how many ounces of green beans are needed to fulfil the demand.

a. Construct a system of three linear equations in three unknowns that represents a mathematical model for this situation.

b. Solve the system from part (a) using matrix methods to determine how much of each of the three items goes into a meal.

Solution a. Suppose that we let M represent the number of ounces of meatloaf needed P represent the number of scoops of mashed potatoes, and G represent the number of ounces of green beans. We can organize the given information in the following table

Meatloaf M

Mashed Potatoes Ρ

Green Beans G

Demand

Calories

75

60

10

475

Protein

7

2

1

36

Carbs

6

10

2

55

This table sets the stage to write the needed system of equations and to get the entries in the matrices we will need to solve the system.

We look at each of the categories—calories, protein, and carbs—separately. The total number of calories in a meal is therefore made up of 75 calories from each of the M ounces of meatloaf, 60 calories from each of the P scoops of mashed potatoes, and 10 calories from each of the G ounces of green beans. That is,

Calories

Equationequals 75 upper M plus 60 upper P plus 10 upper G
.

Similarly, the total number of grams of protein is given by

Protein

Equationequals 7 upper M plus 2 upper P plus 1 upper G
,

and the total number of grams of carbs is given by

Carbs

Equationequals 6 upper M plus 10 upper P plus 2 upper G
.

Since the dietician needs to have a meal with 475 calories, 36 grams of protein, and 55 grams of carbs, these demand values give a system of three equations in three unknowns:

Calories:

Equation75 upper M plus 60 upper P plus 10 upper G equals 475

Protein:

Equation7 upper M plus 2 upper P plus 1 upper G equals 36

Carbs:

Equation6 upper M plus 10 upper P plus 2 upper G equals 55
.

b. For this system, the coefficient matrix A is

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 75 2nd Column 60 3rd Column 10 2nd Row 1st Column 7 2nd Column 2 3rd Column 1 3rd Row 1st Column 6 2nd Column 10 3rd Column 2 EndMatrix

and the matrix of constants (sometimes called the demand matrix ) is

Equationupper B equals Start 3 By 1 Matrix 1st Row 475 2nd Row 36 3rd Row 55 EndMatrix period

The resulting matrix equation is

Equationupper A upper X equals upper B

and the solution is

Equationupper X equals upper A Superscript negative 1 Baseline upper B equals Start 3 By 1 Matrix 1st Row 3.914 2nd Row 2.386 3rd Row 3.829 EndMatrix

or, more meaningfully, the meal should consist of 4 ounces of meatloaf, 2 scoops of mashed potatoes, and 4 ounces of green beans.

This image is of a statue used as a bulletpoint

Thought Experiment

How much effect does rounding the values have on the demand values?

Solving the Regression Equations

Suppose that we have a set of Equationleft-parenthesis x comma y right-parenthesis data that falls into a roughly linear pattern. We can use linear regression to obtain the equation of the line that is the best fit to the data in the sense that it is the one line that comes closest to all the data points, as we discussed in Section 3.4. The process of applying the criterion that the desired line

Equationy equals a x plus b
comes closest to all of the data points when calculating the regression line, however, involves creating and solving a system of two linear equations in the unknown parameters a and b . In particular, it turns out that the two equations are

Equationa dot left-parenthesis sigma-summation x right-parenthesis plus b dot n equals left-parenthesis sigma-summation y right-parenthesis
(1)

Equationa dot left-parenthesis sigma-summation x squared right-parenthesis plus b dot left-parenthesis sigma-summation x right-parenthesis equals left-parenthesis sigma-summation x y right-parenthesis
, (2)

where n is the number of data points and

Equationupper Sigma x
represents the sum of all the x -values,

Equationupper Sigma y
represents the sum of all the y -values,

Equationupper Sigma x y
represents the sum of all the products of the pairs of x - and y -values, and

Equationupper Sigma x squared
represents the sum of the squares of all the x -values.

We discuss summation notation using

Equationupper Sigma
in Appendix A.

Equations (1) and (2) are called the regression equations . We illustrate the process of creating these equations and solving them, using matrices, in the following example. Note that this is not intended as a substitute for using the linear regression features of your calculator or spreadsheet, but rather to demonstrate an important use of matrices and, simultaneously, to provide you with a deeper understanding of some of the issues we raised in the last section.

Example 2 Consider the set of data

X

1

2

3

4

5

6

y

10

22

31

43

54

62

a. Use regression techniques to find the equation of the regression line that fits the data.

b. Create the regression equations in which the coefficients a and b in

Equationy equals a x plus b
are the unknowns as defined in Equations (1) and (2) above.

c. Solve the regression equations from part (b) using matrix methods and compare the results to those from part (a).

Solution a. When we enter these data values into a calculator or spreadsheet and select linear regression, we get

Equationy equals 10.514 x plus 0.2
as the equation of the regression line.

b. To calculate the necessary terms for the regression equations (1) and (2), we rewrite the table of data values in columns and add extra columns for the values of

Equationx squared
and Equationx y.

x

y

Equationx squared

Equationx y

1

10

1

10

2

22

4

44

3

31

9

93

4

43

16

172

5

54

25

270

6

62

36

372

Equationupper Sigma x equals 21

Equationupper Sigma y equals 222

Equationupper Sigma x squared equals 91

Equationupper Sigma x y equals 961

In addition, we also record the sum of all the entries in each column in the bottom row. Thus, the sum of all the x values,

Equationupper Sigma x equals 1 plus 2 plus 3 plus ellipsis plus 6 equals 21
and the sum of all the y values in the second column is
Equationsigma-summation y equals 10 plus 22 plus ellipsis plus 62 equals 222
. In the third column, we enter the square of each of the x values, so that
Equation1 squared equals 1
,
Equation2 squared equals 4
, . . . ,
Equation6 squared equals 36
and the sum of all these values is
Equationupper Sigma x squared equals 91
. In the fourth column, we enter the values for the products of each of the pairs of x - and y -values; thus,
Equation1 times 10 equals 10
,
Equation2 times 22 equals 44
, Equationellipsis ,
Equation6 times 62 equals 372
. The sum of all these products is
Equationsigma-summation x y equals 961
. Finally, note that there are
Equationn equals 6
data points.

We now substitute these values for the sums into the regression equations (1) and (2). Equation (1) becomes

Equationa dot left-parenthesis 21 right-parenthesis plus b dot 6 equals 222

or, equivalently,

Equation21 a plus 6 b equals 222 period

Similarly, Equation (2) becomes

Equationa dot left-parenthesis 91 right-parenthesis plus b dot left-parenthesis 21 right-parenthesis equals 961

or, equivalently,

Equation91 a plus 21 b equals 961 period

c. To solve this system of linear equations using matrices, we introduce the matrix of coefficients and the matrix of constants

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 21 2nd Column 6 2nd Row 1st Column 91 2nd Column 21 EndMatrix
and
Equationupper B equals StartBinomialOrMatrix 222 Choose 961 EndBinomialOrMatrix
,

along with the matrix of unknowns,

Equationupper X equals StartBinomialOrMatrix a Choose b EndBinomialOrMatrix period

Using a calculator, we find that the solution of the matrix equation Equationupper A upper X equals upper B is

Equationupper X equals upper A Superscript negative 1 Baseline upper B equals StartBinomialOrMatrix 10.514 Choose 0.2 EndBinomialOrMatrix equals StartBinomialOrMatrix a Choose b EndBinomialOrMatrix period

That is,

Equationa equals 10.514
and
Equationb equals 0.2
, so that the equation of the regression line is
Equationy equals 10.514 x plus 0.2
, which is identical to the solution we obtained in part (a) using the regression features of a calculator or spreadsheet.

In Section 3.4, we suggested that it often makes sense to scale down large data values to make the computations simpler. Otherwise, it is possible to get overflow errors when the size of the quantities involved exceed the capabilities of the technology being used. Suppose that the independent variable x that we used in Example 2 represents the number of years since 2000, so

Equationx equals 1
corresponds to 2001, etc. We now illustrate what happens if we use the full year instead.

Example 3 Consider the set of data

x

2001

2002

2003

2004

2005

2006

y

10

22

31

43

54

62

a. Find the equation of the regression line using a linear regression routine and compare the coefficients to those in Example 2.

b. Create the regression equations in which the coefficients a and b in

Equationy equals a x plus b
are the unknowns and compare the coefficients in the regression equations to those in Example 2.

c. Solve the regression equations to obtain the equation of the regression line to fit this set of data.

d. Use both the regression line from Example 2 and the regression line from part (b) to predict the next term (when

Equationx equals 2007
here) in the data.

Solution a. When we perform linear regression on this set of data, we get

Equationy equals 10.514 x minus 21 comma 028.371
. In Example 2, the regression line was
Equationy equals 10.514 x plus 0.2
, where we have the same slope, but a dramatically different vertical intercept. If you think about this, it makes sense. The points have the same heights and the same relative positions as those in Example 2, so the slope should be the same. But, by changing the x 's from 1, 2, Equationellipsis , 6 to 2001, 2002, Equationellipsis , 2006, we have essentially moved forward 2000 years, so the vertical intercept is now 2000 years in the past along a line that rises about 10 and a half units each year.

b. As in Example 2, we rewrite the table of data values in columns and add extra columns for the values of

Equationx squared
and Equationx y.

x

y

Equationx squared

Equationx y

2001

10

4004001

20010

2002

22

4008004

44044

2003

31

4012009

62093

2004

43

4016016

86172

2005

54

4020025

108270

2006

62

4024036

124372

Equationupper Sigma x equals 12 comma 021

Equationupper Sigma y equals 222

Equationupper Sigma x squared equals 24 comma 084 comma 091

Equationsigma-summation x y equals 444 comma 961

Notice how large the sums have become compared to the equivalent sums in Example 1. When we substitute these sums into the regression equations, Equation (1) becomes

Equationa dot left-parenthesis 12 comma 021 right-parenthesis plus b dot 6 equals 222

or equivalently,

Equation12 comma 021 a plus 6 b equals 222 period

Similarly, Equation (2) becomes

Equation24 comma 084 comma 091 a plus 12 comma 021 b equals 444 comma 961 period

In this case, the equations are considerably more complicated than those in Example 2 and, worse, the size of the coefficients could potentially lead to the overflow errors mentioned above.

c. To solve the regression equations

Equation12 comma 021 a plus 6 b equals 222
,

Equation24 comma 084 comma 091 a plus 12 comma 021 b equals 444 comma 961
,

we introduce the matrix A of coefficients and the matrix B of constants

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 12 comma 021 2nd Column 6 2nd Row 1st Column 24 comma 084 comma 091 2nd Column 12 comma 021 EndMatrix and Equationupper B equals StartBinomialOrMatrix 222 Choose 444 comma 961 EndBinomialOrMatrix.

Some calculators give the solution of the matrix equation

Equationupper A upper X equals upper B
as

Equationupper X equals StartBinomialOrMatrix a Choose b EndBinomialOrMatrix equals upper A Superscript negative 1 Baseline upper B equals StartBinomialOrMatrix 10.5143 Choose negative 21028.4 EndBinomialOrMatrix
.

This gives the equation of the regression line as

Equationy equals 10.5143 x minus 21 comma 028.4
, which is essentially equivalent to what we found in part (a). However, many other calculator models are unable to handle the large numbers correctly and so only give an error message that suggests that the matrix A does not have an inverse. (We discuss when a matrix does not have an inverse later in the section.)

d. To predict the "next" point

Equationx equals 7
in the data from Example 2, we substitute this into the equation of the regression line
Equationy equals 10.514 x plus 0.2
to get

Equationy equals 10.514 left-parenthesis 7 right-parenthesis plus 0.2 equals 73.798
.

This is a reasonable prediction considering that the value given in the table in Example 2 when

Equationx equals 6
is 62 and the slope of the regression line is slightly more than 10.5. The "next" point in the data set here is
Equationx equals 2007
, and when we substitute it into the equation of the regression line from part (b), we get

Equationy equals 10.5143 left-parenthesis 2007 right-parenthesis minus 21 comma 028.4 equals 73.8001
.

which is fairly close to the value we got with the other equation. (The difference is likely due to rounding the coefficients.)

Clearly, scaling down the values as we did in Example 2 is a much easier, safer, and better approach.

Balancing Chemical Equations

The process of photosynthesis involves the chemical reaction in which carbon dioxide,

Equationupper C upper O 2
, combines with water,
Equationupper H 2 upper O
, to produce glucose,
Equationupper C 6 upper H 12 upper O 6
, plus oxygen,
Equationupper O 2
. That is, several molecules of carbon dioxide,
Equationupper C upper O 2
, (each of which consists of one atom of carbon and two atoms of oxygen) combine with molecules of water,
Equationupper H 2 upper O
, (consisting of two atoms of hydrogen and one of oxygen) to produce molecules of glucose,
Equationupper C 6 upper H 12 upper O 6
, (consisting of six atoms of carbon, 12 atoms of hydrogen, and six atoms of oxygen) and molecules of oxygen,
Equationupper O 2
, (each molecule of free oxygen consists of two atoms of oxygen). This is typical of every chemical reaction, which involves changing one or more substances into one or more different substances by regrouping atoms or ions to form the other substances.

Perhaps the most common problem in elementary chemistry is that of balancing a chemical equation —that is, determining precisely how many molecules of each chemical are needed to make the equation "work". For photosynthesis, the balanced chemical equation turns out to be:

Equation6 dot upper C upper O 2 plus 6 dot upper H 2 upper O equals 6 dot upper O 2 plus upper C 6 upper H 12 upper O 6
;

that is, it takes six molecules of carbon dioxide to combine with six molecules of water to produce one molecule of glucose and six molecules of oxygen. The reason that chemists describe this equation as being in balance is as follows. Consider the number of atoms of each element:

1. There are six atoms of carbon on the left (in

Equationupper C upper O 2
) and six atoms of carbon on the right (in
Equationupper C 6 upper H 12 upper O 6
).

2. There are 12 atoms of hydrogen on the left (two atoms of hydrogen in each of the six molecules of

Equationupper H 2 upper O
) and 12 atoms of hydrogen on the right (in
Equationupper C 6 upper H 12 upper O 6
).

3. There are 18 atoms of oxygen on the left (two in each of the six molecules of carbon dioxide plus six more in the water) and there are 18 atoms of oxygen on the right (two in each of the six molecules of oxygen

Equationupper O 2
and six more in the molecule of
Equationupper C 6 upper H 12 upper O 6
). Because the same number of atoms of each element are on both sides, chemists say that the reaction is balanced.

The problem in chemistry is not looking at a balanced equation and seeing why it is in balance; rather, the problem is to determine the number of molecules of each substance that are needed to achieve balance. In chemistry, this is typically approached using a trial-and-error method, which is usually tedious and often confusing to many students. Alternatively, it turns out to be a simple application of our present methods—just set up a system of linear equations and solve it using matrix methods. We illustrate this systematic approach in several examples.

Example 4 Use matrix methods to balance the chemical equation for the photosynthesis reaction.

Solution In this chemical reaction, we can think of the glucose,

Equationupper C 6 upper H 12 upper O 6
, as our "target" molecule, so that we want to produce one molecule of
Equationupper C 6 upper H 12 upper O 6
We then ask: To produce that one molecule of glucose, how many molecules of carbon dioxide do we need, how many molecules of water do we need, and how many molecules of oxygen are produced as a by-product of the reaction? Let x be the number of molecules of
Equationupper C upper O 2
, let y be the number of molecules of
Equationupper H 2 upper O
, and let z be the number of molecules of
Equationupper O Subscript 2
. We can then write this reaction as

Equationx dot upper C upper O 2 plus y dot upper H 2 upper O right-arrow upper C 6 upper H 12 upper O 6 plus z dot upper O 2
.

Equivalently, we can think of this reaction equation as

Equationx dot upper C upper O 2 plus y dot upper H 2 upper O minus z dot upper O 2 right-arrow upper C 6 upper H 12 upper O 6
,

where we have isolated our target molecule

Equationupper C 6 upper H 12 upper O 6
on the right-hand side of the reaction equation. We now count the number of atoms of each of the elements separately.

Carbon:

Equation1 x plus 0 y minus 0 z equals 6

Hydrogen:

Equation0 x minus 2 y minus 0 z equals 12

Oxygen:

Equation2 x plus 1 y minus 2 z equals 6 period

This system of three linear equations in three unknowns is particularly simple and can actually be solved very easily using algebra. The first equation reduces to

Equationx equals 6
and the second equation reduces to
Equation2 y equals 12
, so that
Equationy equals 6
. Substituting these values into the third equation, for oxygen, we find that

Equation2 left-parenthesis 6 right-parenthesis plus 1 left-parenthesis 6 right-parenthesis minus 2 z equals 6

or equivalently

Equation12 plus 6 minus 2 z equals 6
.

If we add 2z to both sides of the equation and subtract 6 from both sides of the equation, we have

Equation12 equals 2 z
,

so that

Equationz equals 6
.

Thus, our solution gives

Equation6 dot upper C upper O 2 plus 6 dot upper H 2 upper O right-arrow 6 dot upper O 2 plus upper C 6 upper H 12 upper O 6
,

as before.

Alternatively, we can solve this system of equations using matrix methods. We can write the system as the matrix equation

Equationupper A upper X equals upper B
, where the coefficient matrix and the matrix of constants are

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column 2 3rd Column 0 3rd Row 1st Column 2 2nd Column 1 3rd Column negative 2 EndMatrix and Equationupper B equals Start 3 By 1 Matrix 1st Row 6 2nd Row 12 3rd Row 6 EndMatrix.

The solution is

Equationupper X equals upper A Superscript negative 1 Baseline upper B equals Start 3 By 1 Matrix 1st Row 6 2nd Row 6 3rd Row 6 EndMatrix

Therefore,

Equationx equals 6
,
Equationy equals 6
, and
Equationz equals 6
, so that the balanced equation is

Equation6 dot upper C upper O 2 plus 6 dot upper H 2 upper O minus 6 dot upper O 2 right-arrow upper C 6 upper H 12 upper O 6

or, in more standard format,

Equation6 dot upper C upper O 2 plus 6 dot upper H 2 upper O right-arrow upper C 6 upper H 12 upper O 6 plus 6 dot upper O 2
,

as before.

Example 5 The main constituent of natural gas is methane. When natural gas burns in air, the methane combines with oxygen to produce carbon dioxide and water. That is, molecules of methane

Equationupper C upper H 4
(which consist of one atom of carbon and four atoms of hydrogen) combine with molecules of oxygen
Equationupper O 2
(which consist of two atoms of oxygen) to produce molecules of carbon dioxide,
Equationupper C upper O 2
(which consist of one atom of carbon and two of oxygen) and molecules of water
Equationupper H 2 upper O
(which consist of two atoms of hydrogen and one of oxygen). Use matrix methods to balance the chemical equation for the reaction when methane burns if:

a. the target is

Equationupper C upper O 2
;

b. the target is

Equationupper H 2 upper O
.

Solution a. To produce one molecule of carbon dioxide

Equationupper C upper O 2
as our target, we ask: how many molecules of methane do we need, how many molecules of oxygen do we need, and, how many molecules of water are produced as a by-product of the reaction? Let x be the number of molecules of
Equationupper C upper H 4
, let y be the number of molecules of
Equationupper O 2
, and let z be the number of molecules of
Equationupper H 2 upper O
. We can write this reaction as

Equationx dot upper C upper H 4 plus y dot upper O 2 right-arrow upper C upper O 2 plus z dot upper H 2 upper O
.

Equivalently, we can think of this reaction equation as

Equationx dot upper C upper H 4 plus y dot upper O 2 minus z dot upper H 2 upper O right-arrow upper C upper O 2
,

where we have isolated our target molecule

Equationupper C upper O 2
on the right-hand side of the reaction equation. We now count the number of atoms of each of the elements separately.

Carbon:

Equation1 x plus 0 y minus 0 z equals 1

Hydrogen:

Equation4 x plus 0 y minus 2 z equals 0

Oxygen:

Equation0 x plus 2 y minus 1 z equals 2 period

This system of three linear equations in three unknowns can be written as the matrix equation

Equationupper A upper X equals upper B
, where the coefficient matrix A and the matrix of constants B are

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 2nd Row 1st Column 4 2nd Column 0 3rd Column negative 2 3rd Row 1st Column 0 2nd Column 2 3rd Column negative 1 EndMatrix and Equationupper B equals Start 3 By 1 Matrix 1st Row 1 2nd Row 0 3rd Row 2 EndMatrix.

The solution is

Equationupper X equals upper A Superscript negative 1 Baseline upper B
, which is

Equationupper X equals upper A Superscript negative 1 Baseline upper B equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix equals Start 3 By 1 Matrix 1st Row 1 2nd Row 2 3rd Row 2 EndMatrix.

Since

Equationx equals 1
,
Equationy equals 2
, and
Equationz equals 2
, the balanced equation is

Equation1 dot upper C upper H 4 plus 2 dot upper O 2 minus 2 dot upper H 2 upper O right-arrow upper C upper O 2

or, in more standard format,

Equationupper C upper H 4 plus 2 dot upper O 2 right-arrow upper C upper O 2 plus 2 dot upper H 2 upper O
.

b. To produce one molecule of water,

Equationupper H 2 upper O
, as our target, we ask: How many molecules of methane do we need, how many molecules of oxygen do we need, and how many molecules of carbon dioxide are produced as a by-product of the reaction? Let x be the number of molecules of
Equationupper C upper H 4
, let y be the number of molecules of
Equationupper O 2
, and let z be the number of molecules of
Equationupper C upper O 2
. So we can write this reaction as

Equationx dot upper C upper H 4 plus y dot upper O 2 right-arrow z dot upper C upper O 2 plus upper H 2 upper O

or, equivalently,

Equationx dot upper C upper H 4 plus y dot upper O 2 minus z dot upper C upper O 2 right-arrow upper H 2 upper O
,

where now the target molecule

Equationupper H 2 upper O
is alone on the right-hand side. We again count the number of atoms of each of the elements separately.

Carbon:

Equation1 x plus 0 y minus 1 z equals 0

Hydrogen:

Equation4 x plus 0 y minus 0 z equals 2

Oxygen:

Equation0 x plus 2 y minus 2 z equals 1 period

This new system of equations can also be written in matrix form as

Equationupper A upper X equals upper B
and its solution is

Equationupper X equals upper A Superscript negative 1 Baseline upper B equals Start 3 By 1 Matrix 1st Row 0.5 2nd Row 1 3rd Row 0.5 EndMatrix equals Start 3 By 1 Matrix 1st Row one-half 2nd Row 1 3rd Row one-half EndMatrix
.

Therefore, the balanced equation is

Equationone-half dot upper C upper H 4 plus 1 dot upper O 2 minus one-half dot upper C upper O 2 right-arrow upper H 2 upper O
,

or equivalently

Equationone-half dot upper C upper H 4 plus 1 dot upper O 2 right-arrow one-half dot upper C upper O 2 right-arrow upper H 2 upper O
.

Because it makes little sense to have halves of molecules, we multiply both sides of this equation by 2 to eliminate the fractions and end up with the balanced equation

Equationupper C upper H 4 plus 2 dot upper O 2 right-arrow upper C upper O 2 plus 2 dot upper H 2 upper O
,

which is the same equation we had before.

In this example, we looked at one of the products of a reaction as our "target". Alternatively, we can think of one of the original reactants as our "target" in the process of balancing a chemical reaction. We illustrate this in the following example.

Example 6 When dimethyl ether,

Equationupper C 2 upper H 6 upper O
, burns in oxygen,
Equationupper O 2
, it produces carbon dioxide,
Equationupper C upper O 2
, and water,
Equationupper H 2 upper O
. Balance the chemical reaction equation for this reaction.

Solution Suppose we have one molecule of dimethyl ether,

Equationupper C 2 upper H 6 upper O
, and think of it as our target. We then ask: To use up that one molecule of dimethyl ether, how many molecules of oxygen do we need how many molecules of carbon dioxide will be produced as a product of the reaction, and how many molecules of water are produced as a by-product of the reaction? Let x be the number of molecules of
Equationupper O 2
, let y be the number of molecules of
Equationupper C upper O 2
, and let z be the number of molecules of
Equationupper H 2 upper O
. We can write this reaction as

Equationupper C 2 upper H 6 upper O plus x dot upper O 2 right-arrow y dot upper C upper O 2 plus z dot upper H 2 upper O
.

Equivalently, we can think of this reaction equation as

Equationupper C 2 upper H 6 upper O right-arrow negative x dot upper O 2 right-arrow y dot upper C upper O 2 plus z dot upper H 2 upper O
,

where we have isolated our target molecule

Equationupper C 2 upper H 6 upper O
on the left-hand side of the reaction equation. We now count the number of atoms of each of the elements separately.

Carbon:

Equation2 equals minus 0 x plus 1 y plus 0 z

Hydrogen:

Equation6 equals minus 0 x plus 0 y plus 2 z

Oxygen:

Equation1 equals minus 2 x plus 2 y plus 1 z
.

We write this system of three linear equations in three unknowns as the matrix equation

Equationupper A upper X equals upper B
, where the coefficient matrix and the matrix of constants are

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 0 2nd Column 1 3rd Column 0 2nd Row 1st Column 0 2nd Column 0 3rd Column 2 3rd Row 1st Column negative 2 2nd Column 2 3rd Column 1 EndMatrix
and
Equationupper A equals Start 3 By 1 Matrix 1st Row 2 2nd Row 6 3rd Row 1 EndMatrix
.

The solution

Equationupper X equals upper A Superscript negative 1 Baseline upper B
is therefore

Equationupper X equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix equals Start 3 By 1 Matrix 1st Row 3 2nd Row 2 3rd Row 3 EndMatrix

and so the balanced equation is

Equationupper C 2 upper H 6 upper O plus 3 dot upper O 2 right-arrow 2 dot upper C upper O 2 plus 3 dot upper H 2 upper O
.

We note that the system of equations that arose in this last example is actually very simple to solve by hand because so many of the coefficients are 0. In particular, the system of equations

Equationminus 0 x plus 1 y plus 0 z equals 2

Equationminus 0 x plus 0 y plus 2 z equals 6

Equationminus 2 x plus 2 y plus 1 z equals 1

is actually equivalent to

Equationy equals 2

Equation2 z equals 6

Equationminus 2 x plus 2 y plus z equals 1

The first two equations tell us that

Equationy equals 2
and
Equationz equals 3
. Therefore, when we substitute into the third equation, we find that

Equationminus 2 x plus 2 y plus z equals minus 2 x plus 2 left-parenthesis 2 right-parenthesis plus 3 equals 1

or

Equationminus 2 x plus 4 plus 3 equals 1
.

If we add

Equation2 x
to both sides of this equation and subtract 1 from both sides, we have

Equation2 x equals 4 plus 3 minus 1 equals 6
,

and therefore

Equationx equals 3
, which is precisely the same solution we found using matrix methods.

Not Every System of Equations has a Solution

We have said that the solution of a system of linear equations in matrix form

Equationupper A upper X equals upper B
has the solution
Equationupper X equals upper A Superscript negative 1 Baseline upper B
. However, there is a catch—not every square matrix A has an inverse
Equationupper A Superscript negative 1
. (We will discuss what this means in the next section.) In such a case, either there is no solution to the system of linear equations or there are infinitely many solutions. For example, consider the system of linear equations

Equation2 x plus y equals 5

Equation2 x plus y equals 8

This system of equations does not have a solution; that is, there is no pair of values for x and y that can possibly satisfy both of these equations simultaneously. To see why, let's think about this system of equations graphically. Each equation represents a line in the plane. Suppose we write each of the equations in slope-intercept form. For the first equation, we solve for y and so obtain

Equationy equals minus 2 x plus 5
.

Similarly, when solving for y , the second equation becomes

Equationy equals minus 2 x plus 8
.

Therefore, we see that both lines have slope of

Equationnegative 2
. Since they have the same slope, they are parallel, as seen in Figure 4.4, and consequently the lines never intersect. So, there cannot be a solution.

How does this play out with matrices? The corresponding coefficient matrix is

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 1 2nd Row 1st Column 2 2nd Column 1 EndMatrix
.

If you enter

Equationupper A Superscript negative 1
for the inverse of this matrix on your calculator, it will respond with an error message about a "singular matrix". A singular matrix is a matrix that does not have an inverse. Fortunately, most square matrices do have inverses, so that the corresponding system of linear equations does have a solution (picture how unlikely it is for two lines in the plane to be parallel).

Figure 4.4

This image shows two linear equations on a blank graph. The first line is labeled as y Equals Minus 2 x Plus 8. The second line is labelled y Equals Minus 2 x Plus 5. The lines start and end in Quadrants  2 and 4 while passing through 1.

The same ideas apply to systems of more than two equations in two unknowns, although it is more difficult to give a geometric interpretation to a system of more than three linear equations. Usually, most square matrices will have an inverse and the matrix expression

Equationupper X equals upper A Superscript negative 1 Baseline upper Beta
will make sense and give you the solution; in the rare cases when there is no solution, you will get the error message about a singular matrix.

Some Systems of Equations have Multiple Solutions

Another potential complication occurs when a system of linear equations has infinitely many different solutions, although it is also a fairly rare situation. Consider the system of equations

Equation2 x plus y equals 5

Equation4 x plus 2 y equals 10

If you examine these two equations carefully, you will notice that the second equation is precisely twice the first. So the second equation provides no additional information about the two variables x and y . Effectively we only have one equation in the two unknowns. This is not enough information to solve for the two variables, so we cannot get a single, or unique, solution. For instance,

Equationx equals 1
and
Equationy equals 3
is one solution to the system. However,
Equationx equals 2
and
Equationy equals 1
is also a solution, and for that matter, so is
Equationx equals 0
and
Equationy equals 5
, as is
Equationx equals 3
and
Equationy equals negative 1
. There are, in fact, infinitely many solutions. Geometrically, the two lines are identical, so they "meet" at every point and there are infinitely many common points.

To see what happens algebraically, the only equation we have is

Equation2 x plus y equals 5
. Therefore, we can isolate y as
Equationy equals 5 minus 2 x
and for every possible value of x , there is a corresponding value for y . Each of these pairs Equationleft-parenthesis x comma y right-parenthesis is then a solution to the system.

From a matrix point of view, the corresponding coefficient matrix is also singular and so it does not have an inverse. Again, fortunately, this is a very rare situation and usually arises fairly infrequently in practice. However, it does occur occasionally in the process of balancing chemical reactions, so we consider what happens and how to circumvent the difficulty in the following examples.

Example 7 When aluminum, Al, is treated with sulfuric acid

Equationupper H 2 upper S upper O 4
, the result is aluminum sulfate,
Equationupper A l 2 left-parenthesis upper S upper O 4 right-parenthesis Subscript 3 Baseline
, (which has three
Equationupper S upper O 4
ions) with free hydrogen,
Equationupper H 2
, as a by-product. Balance the chemical reaction equation for this reaction.

Solution Suppose we want to produce one molecule of aluminum sulfate,

Equationupper A l 2 left-parenthesis upper S upper O 4 right-parenthesis Subscript 3
, which we think of as our target. We then ask: To produce that one molecule of aluminum sulfate, how many molecules of aluminum do we need, how many molecules of sulfuric acid do we need, and how many molecules of hydrogen are produced as a by-product of the reaction? Let x be the number of molecules of Equationupper A l, let y be the number of molecules of
Equationupper H 2 upper S upper O 4
, and let z be the number of molecules of
Equationupper H 2
. We can write this reaction as

Equationx dot upper A l plus y dot upper H 2 upper S upper O 4 right-arrow upper A l 2 left-parenthesis upper S upper O 4 right-parenthesis Subscript 3 Baseline plus z dot upper H 2
.

Equivalently, we can think of this reaction equation as

Equationx dot upper A l plus y dot upper H 2 upper S upper O 4 minus z dot upper H 2 right-arrow upper A l 2 left-parenthesis upper S upper O 4 right-parenthesis Subscript 3 Baseline
,

where we have isolated our target molecule

Equationupper A l 2 left-parenthesis upper S upper O 4 right-parenthesis Subscript 3 Baseline
on the right-hand side of the reaction equation. We now count the number of atoms of each of the elements separately.

Aluminum:

Equation1 x plus 0 y minus 0 z equals 2

Hydrogen:

Equation0 x plus 2 y minus 2 z equals 0

Sulfur:

Equation0 x plus 1 y minus 0 z equals 3

Oxygen:

Equation0 x plus 4 y minus 0 z equals 12
.

Notice that this is actually a system of four linear equations in the three unknowns x , y , and z .

This system is simple enough that we can approach it algebraically. The first equation immediately tells us that

Equationx equals 2
and the third equation gives
Equationy equals 3
. Substituting these values into the second equation, we have

Equation2 y minus 2 z equals 0

Equation2 dot 3 minus 2 z equals 0

Equation2 z equals 6

or

Equationz equals 3
.

The three values

Equationx equals 2
,
Equationy equals 3
and
Equationz equals 3
also satisfy the fourth equation since

Equation0 dot 2 plus 4 dot 3 minus 0 dot 3 equals 12
.

If you look carefully at the four equations, you will notice that the fourth equation is precisely four times the third, so the last equation does not provide any new information. What this means is that only the first three equations give useful information, so that we really have only three equations in the three unknowns.

Notice that, if we tried to use matrix methods unthinkingly, we would take as the coefficient matrix

Equationupper A equals Start 4 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column 2 3rd Column negative 2 3rd Row 1st Column 0 2nd Column 1 3rd Column 0 4th Row 1st Column 0 2nd Column 4 3rd Column 0 EndMatrix
.

Since only a square matrix may have an inverse, there is obviously no inverse

Equationupper A Superscript negative 1
for this matrix and so we could not proceed.

Because we noticed that only the first three equations give useful information and the fourth equation is simply a multiple of the third equation, we can write this system of three linear equations in three unknowns in matrix form as

Equationupper A upper X equals upper B
, where the coefficient matrix and the matrix of constants are

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column 2 3rd Column negative 2 3rd Row 1st Column 0 2nd Column 1 3rd Column 0 EndMatrix
and
Equationupper B equals Start 3 By 1 Matrix 1st Row 2 2nd Row 0 3rd Row 3 EndMatrix
.

The solution

Equationupper X equals upper A Superscript negative 1 Baseline upper B
is therefore

Equationupper X equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix equals Start 3 By 1 Matrix 1st Row 2 2nd Row 3 3rd Row 3 EndMatrix

and so the balanced equation is

Equation2 dot upper A l plus 3 dot upper H 2 upper S upper O 4 right-arrow upper A 1 Subscript 2 Baseline left-parenthesis upper S upper O 4 right-parenthesis Subscript 3 Baseline plus 3 dot upper H 2
.

The system of equations in Example 7 was simple in the sense that it was fairly evident that the fourth equation was four times the third equation. Things are not usually quite that simple, as we illustrate in the next example.

Example 8 Sulfuric acid

Equationupper H 2 upper S upper O Baseline 4
, and sodium hydroxide,
Equationupper N a upper O upper H
, combine to produce sodium sulfate,
Equationupper N a 2 upper S upper O 4
, with water,
Equationupper H 2 upper O
, as the by-product. Find the balanced chemical reaction equation for this reaction.

Solution We will think of sodium sulfate,

Equationupper N a 2 upper S upper O 4
, as our target and so ask: To produce one molecule of sodium sulfate, how many molecules of sulfuric acid do we need how many molecules of sodium hydroxide do we need and how many molecules of water are produced as a by-product of the reaction? Let x be the number of molecules of
Equationupper H 2 upper S upper O Baseline 4
, let y be the number of molecules of
Equationupper N a upper O upper H
, and let z be the number of molecules of
Equationupper H 2 upper O
. We can write this reaction as

Equationx dot upper H Subscript 2 Baseline SO Subscript 4 Baseline plus y dot NaOH right-arrow Na Subscript 2 Baseline SO Subscript 4 Baseline plus z dot upper H Subscript 2 Baseline upper O
,

which is equivalent to

Equationx dot upper H 2 upper S upper O 4 plus y dot upper N a upper O upper H minus z dot upper H 2 upper O right-arrow upper N a 2 upper S upper O 4
,

where we have isolated our target molecule

Equationupper N a 2 upper S upper O 4
on the right-hand side. We now count the number of atoms of each of the elements separately.

Hydrogen:

Equation2 x plus y minus 2 z equals 0

Sulfur (

Equationupper S
)
Equationx plus 0 y minus 0 z equals 1

Oxygen:

Equation4 x plus y minus z equals 4

Sodium (

Equationupper N a
);
Equation0 x plus y minus 0 z equals 2
.

As in Example 7, this is a system of four linear equations in the three unknowns. Fortunately, this system can be solved easily algebraically. The second equation gives

Equationx equals 1
and the fourth equation gives
Equationy equals 2
. Substituting these values into the first equation, we have

Equation2 dot 1 plus 2 minus 2 z equals 0
,

so that

Equation4 equals 2 z

and so

Equationz equals 2
.

We now check that this solution

Equationx equals 1
,
Equationy equals 2
, and
Equationz equals 2
also satisfies the third equation:

Equation4 x plus y minus z equals 4

Equation4 dot 1 plus 2 minus 2 equals 4 plus 2 minus 2 equals 4
.

So our solution indeed satisfies all four equations and the balanced chemical reaction is

Equationupper H 2 upper S upper O 4 plus 2 dot upper N a upper O upper H right-arrow upper N a 2 upper S upper O 4 plus 2 dot upper H 2 upper O
.

Unlike Example 7, though, none of the four equations is just a simple multiple of one of the other equations. Nevertheless, one of the four equations must be some combination of the other three in the sense that if we add the correct multiples of the first three equations, say, we would get the fourth equation. The key fact is that, with the three unknowns, we only need three pieces of information (3 equations) to solve the system (assuming that it has a solution). However, we then must check to be sure that the solution satisfies the remaining equation. The problem then becomes trying to decide which of the equations is redundant—meaning which is a combination of the other equations and so provides no new information. It is always possible to determine, algebraically, not only which equation can be written in terms of the others, but also what that relationship is. However, this is a somewhat more complicated process and we will not go into it here.

Problems

1. The accompanying table shows the life expectancy of a male or female child born in various years in the U.S.

Male

Female

1970

67.1

74.7

1975

68.8

76.6

1980

70.0

77.4

1985

71.1

78.2

1990

71.8

78.8

1995

72.5

78.9

2000

74.3

79.7

2005

75.2

80.4

Source: 2009 Statistical Abstracts of the U.S.

a. Find the regression line that fits each set of values as a linear function of time t in years since 1970.

b. Use the lines from part (a) to generate a system of two linear equations in two unknowns. Estimate graphically the year in which the life expectancy of males and females will be the same.

c. Solve the system of equations from part (b) algebraically, correct to two decimal place accuracy.

d. Solve the system of equations from part (b) using matrices.

e. Discuss the reasonableness of extending the two linear trends far enough for them to intersect.

2. The accompanying table shows the world record times, in seconds, for the 500 meter freestyle swimming race for men and for women in various years.

1992

1993

1994

1998

1999

2000

2001

2004

2006

2007

2008

Men

21.50

21.31

21.21

21.13

21.10

20.98

20.93

20.48

Women

24.75

24.23

24.09

23.59

a. Find the regression line that fits each set of values as a linear function of time t in years since 1990.

b. Use the lines from part (a) to generate a system of two linear equations in two unknowns. Estimate graphically the year in which the world record times for men and women will be the same.

c. Solve the system of equations from part (b) using matrices.

d. Discuss the reasonableness of extending the two linear trends far enough for them to intersect.

3. The accompanying table shows the world record times, in minutes: seconds, for the 1000 meter speed skating event for men and for women in various years.

1996

1997

1998

1999

2000

2001

2002

2005

2006

2007

Men

Equation1 colon 11.67

Equation1 colon 10.16

Equation1 colon 09.60

Equation1 colon 08.55

Equation1 colon 08.35

Equation1 colon 07.72

Equation1 colon 07.18

Equation1 colon 07.03

Equation1 colon 07.00

Women

Equation1 colon 15.43

Equation1 colon 14.96

Equation1 colon 14.61

Equation1 colon 14.13

Equation1 colon 13.83

Equation1 colon 13.11

a. Find the regression line that fits each set of values as a linear function of time t in years since 1995. (Hint: The times listed are in the minute: second format, so

Equation1 colon 11.67
represents 1 minute and 11.67 seconds. First transform these times to decimal form by dividing the seconds by 60.)

b. Use the lines from part (a) to generate a system of two linear equations in two unknowns. Estimate graphically the year in which the world record times for men and women will be the same.

c. Solve the system of equations from part (b) using matrices.

d. Discuss the reasonableness of extending the two linear trends far enough for them to intersect.

4. When aluminum, Equationupper A l, is treated with hydrochloric acid, Equationupper H upper C l, (which consists of one atom of hydrogen

Equationupper H
and one atom of chlorine
Equationupper C l
), it produces aluminum chloride,
Equationupper A l upper C l 3
, (consisting of one atom of Equationupper A l and 3 atoms of chlorine) and free hydrogen,
Equationupper H 2
. Balance the chemical reaction equation for this reaction.

5. When iron, Equationupper F e, is exposed to oxygen,

Equationupper O 2
, it rusts, producing iron oxide,
Equationupper F e 3 upper O 4
, known as magnetite. Balance the chemical reaction equation for this reaction.

6. When lye, or sodium hydroxide,

Equationupper N a upper O upper H
, (consisting of one atom of sodium, one of oxygen, and one of hydrogen) is mixed with sulfur dioxide,
Equationupper S upper O 2
, (which consists of one atom of sodium Equationupper S and two atoms of oxygen), it produces sodium sulfite,
Equationupper N a 2 upper S upper O 3
, (consisting of two atoms of sodium Equationupper N a, one atom of sulfur Equationupper S, and three atoms of oxygen) and water,
Equationupper H 2 upper O
. Balance the chemical reaction equation for this reaction.

7. Pure alcohol, or ethanol,

Equationupper C 2 upper H 5 upper O upper H
, consists of two atoms of carbon, six atoms of hydrogen (one of the six is paired with oxygen to make what is called an hydroxide ion Equationupper O upper H), and one atom of oxygen. When alcohol is mixed with oxygen,
Equationupper O 2
, it produces carbon dioxide,
Equationupper C upper O 2
, and water,
Equationupper H 2 upper O
. Balance the chemical reaction equation for this reaction.

8. Suppose that the student athletes at Brookdale College need a meal with 700 calories, 40 grams of protein, and 75 grams of carbs. As in Example 1, each ounce of meatloaf has 75 calories, 7 grams of protein, and 6 grams of carbs. Each scoop of mashed potatoes has 60 calories, 2 grams of protein, and 10 grams of carbs. Each ounce of green beans has 10 calories, 1 gram of protein, and 2 grams of carbs. How many ounces of meatloaf, how many scoops of mashed potatoes, and how many ounces of green beans are needed to fulfil this demand?

9. Suppose that the dietician at Brookdale College wants to provide a different option for students who are trying to lose weight, so a meal is to include 375 calories, 30 grams of protein, and 50 grams of carbs. As in Example 1, each ounce of meatloaf has 75 calories, 7 grams of protein, and 6 grams of carbs. Each scoop of mashed potatoes has 60 calories, 2 grams of protein, and 10 grams of carbs. Each ounce of green beans has 10 calories, 1 gram of protein, and 2 grams of carbs. How many ounces of meatloaf, how many scoops of mashed potatoes, and how many ounces of green beans are needed to fulfil this demand?

10. a. Given the set of data Equationleft-parenthesis 1 comma 11 right-parenthesis,Equationleft-parenthesis 2 comma 25 right-parenthesis,Equationleft-parenthesis 3 comma 33 right-parenthesis,Equationleft-parenthesis 4 comma 45 right-parenthesis,Equationleft-parenthesis 5 comma 57 right-parenthesis, and Equationleft-parenthesis 6 comma 65 right-parenthesis, find the equation of the regression line using your calculator or an appropriate software package.

b. Calculate the sums needed for the two regression equations by completing the entries in the following table.

x

y

Equationx squared

Equationx y

1

11

2

25

3

33

4

45

5

57

6

65

Equationupper Sigma x equals

Equationupper Sigma y equals

Equationupper Sigma x squared equals

Equationupper Sigma x y equals

c. Use the results of part (b) to write the system of linear equations in a and b that can be used to determine the values for the unknown coefficients a and b in the regression equation.

d. Solve the system of equations from part (c). How do your results compare to what you obtained directly in part (a)?

e. In Sections 3.4 and 4.1, we suggested that you scale down large numbers, such as the full year 2000, 2001, 2002,Equationellipsis in a set of data. Based on your calculations in part (b), can you explain why this kind of scaling is desirable? In particular, what do you think might happen to the sums if many data values, in the x 's, say, consisted of full years and if the y 's were also large numbers?

11. Repeat parts (a)−(d) of Problem 10 for the data Equationleft-parenthesis 0 comma 24 right-parenthesis,Equationleft-parenthesis 5 comma 21 right-parenthesis,Equationleft-parenthesis 10 comma 17 right-parenthesis,Equationleft-parenthesis 15 comma 14 right-parenthesis,Equationleft-parenthesis 20 comma 12 right-parenthesis, andEquationleft-parenthesis 25 comma 9 right-parenthesis

4.3 Matrix Products and their Applications

In the last two sections, we repeatedly encountered the matrix equation

Equationupper A upper X equals upper Beta
with solution
Equationupper X equals upper A Superscript negative 1 Baseline upper B
, assuming that the inverse matrix
Equationupper A Superscript negative 1
exists. Each of these matrix equations,

Equationupper A upper X equals upper B
and
Equationupper X equals upper A Superscript negative 1 Baseline upper B
,

involves the product of a matrix (either A or

Equationupper A Superscript negative 1
) and a vector (either X or B) and the result of the product is another vector (either B or X, respectively). In each case, the matrix is a square matrix of size n by n , and the vector that it multiplies is a column vector of size n by 1. Note that the number of columns in the matrix matches the number of rows in the vector. In these products, the matrix always multiplies the column vector, so that the matrix always comes to the left of the vector for the product to make sense.

Figure 4.5

This image shows a line that is labeled X Equals Start 2 by 1 Matrix left bracket Row 2 Row 3 right bracket. The line starts at the origin and makes its way until it hits left paren 2 comma 3 right paren as a point.

Figure 4.6

This image shows a line that is labeled X Equals Start 2 by 1 Matrix left bracket Row x Row y right bracket. The line starts at the origin and makes its way until it hits left paren x comma y right paren as a point.

We now investigate what the product of a matrix and a column vector really means by looking at things graphically. First, any vector with two entries can be pictured as a line segment starting at the origin. Thus, if the vector is

Equationupper X equals StartBinomialOrMatrix 2 Choose 3 EndBinomialOrMatrix
, then we picture it as the line segment starting at the origin and ending at the point Equationleft-parenthesis 2 comma 3 right-parenthesis, as shown in Figure 4.5. That is, the entries in the vector become the coordinates of the end point of the line segment.

In general, as shown in Figure 4.6, the vector

Equationupper X equals StartBinomialOrMatrix x Choose y EndBinomialOrMatrix
, whose entries are x and y , can be pictured as the line segment from the origin to the point Equationleft-parenthesis x comma y right-parenthesis. A similar graphical interpretation can be given to vectors with three or more entries. For instance, a vector with 3 entries, x , y , and z (they are also called components ) is pictured as the line segment in three dimensions that starts at the origin (the point where three mutually perpendicular axes meet) and ends at the point in 3-dimensional space whose coordinates are Equationleft-parenthesis x comma y comma z right-parenthesis. In Figure 4.7a, we show the location of the point P in space having coordinates Equationleft-parenthesis x comma y comma z right-parenthesis. Picture a rectangular box with one corner at the origin and with edges lying along the x , the y , and the z axes. We measure a distance x along the x -axis, a distance y along the y -axis, and a distance z along the z -axis to determine the opposite corner P of the box, as shown in Figure 4.7a. The line segment from the origin to this point, as shown in Figure 4.7b, is the three-dimensional image of the vector.

Figure 4.7

This image shows two rectangular boxes with one corner being at the origin. Along the edges of the origin is X and Y showing the horizontal and vertical axes respectively. The z axis goes vertically from the origin to determine the opposite corner. From the origin there is a line that points to point P which is at the top height of the box on the z axis.

When we multiply a vector X and a matrix A, as in

Equationupper A upper X equals upper Y
, the result is a new vector Y. As a result, we can think of the product Equationupper A upper X of a matrix and a vector as the process of transforming one vector, X, into another vector, Y. The matrix A plays the role of the operator that performs the transformation.

Example 1 Given the matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 5 2nd Column negative 1 2nd Row 1st Column 4 2nd Column 3 EndMatrix
,

use a calculator to find the product of the matrix A and the vector

Equationupper X equals StartBinomialOrMatrix 2 Choose 3 EndBinomialOrMatrix
and then draw the graphical image of the two vectors.

Solution The vector X is shown in Figure 4.8; it extends from the origin to the point Equationleft-parenthesis 2 comma 3 right-parenthesis. Using the matrix features of a calculator, we form the product AX and so find that the new column vector

Equationupper Y equals upper A upper X equals StartBinomialOrMatrix 7 Choose 17 EndBinomialOrMatrix
. Notice that Y, which is also shown in Figure 4.8, has a different length from X (it is much longer) and is inclined at a different angle from that of X.

Figure 4.8

This image shows two lines originating from the origin of the graph. The first graph is labeled X and points to a plotted point of left paren 2 comma 3 right paren. The second line is labeled Y Equals A X and points to the plotted point left paren 7 comma 17 right paren.

The length, or magnitude , of a vector X is the distance from the origin to the end of the vector. We write this as

Equationdouble-vertical-bar upper X double-vertical-bar
and we use the usual formula for the distance d between two points
Equationleft-parenthesis x 1 comma y 1 right-parenthesis
and
Equationleft-parenthesis x 2 comma y 2 right-parenthesis
in the plane,

Equationd equals StartRoot left-parenthesis x 2 minus x 1 right-parenthesis squared plus left-parenthesis y 2 minus y 1 right-parenthesis squared EndRoot
,

to calculate the magnitude of a vector. For instance, the vector

Equationupper X equals StartBinomialOrMatrix 2 Choose 3 EndBinomialOrMatrix
extends from the origin Equationleft-parenthesis 0 comma 0 right-parenthesis to the point Equationleft-parenthesis 2 comma 3 right-parenthesis, as shown in Figure 4.9, so the distance formula gives its magnitude as

Equationdouble-vertical-bar upper X double-vertical-bar equals StartRoot left-parenthesis 2 minus 0 right-parenthesis squared plus left-parenthesis 3 minus 0 right-parenthesis squared EndRoot equals StartRoot 4 plus 9 EndRoot equals StartRoot 13 EndRoot almost-equals 3.6

Similarly, the magnitude of the vector

Equationupper Y equals StartBinomialOrMatrix 7 Choose 17 EndBinomialOrMatrix
is

Equationdouble-vertical-bar upper Y double-vertical-bar equals StartRoot 7 squared plus 17 squared EndRoot equals StartRoot 49 plus 289 EndRoot equals StartRoot 338 EndRoot almost-equals 18.38

so this vector is slightly more than 5 times as long as X.

In general, the magnitude of any vector

Equationupper X equals StartBinomialOrMatrix x Choose y EndBinomialOrMatrix
is

Equationdouble-vertical-bar upper X double-vertical-bar equals StartRoot x squared plus y squared EndRoot
,

as illustrated in Figure 4.10.

Figure 4.9

This image shows an equation that originates from the origin of the graph. The equation is labeled as Double Absolute value of X Equals Square Root 2 cubed Plus 3 Cubed. The line points to the plotted point of left paren 2 comma 3 right paren.

Figure 4.10

This image shows an equation that originates from the origin of the graph. The equation is labeled as Double Absolute value of X Equals Square Root x cubed Plus y Cubed. The line points to the plotted point of left paren x comma y right paren.

Two Competing Populations

We next consider how the product of a matrix and a vector can be used to model the populations of two competing species over time. Suppose that herds of sheep and cattle are competing for the same environment, say the grass in an isolated valley in the mountains, and we wish to model the two populations over time. Initially, at time

Equationt equals 0
, there are
Equationupper S 0
sheep and
Equationupper C 0
cattle. Suppose that, if there were no sheep competing for the grass, the cattle population would grow by 40% a year, so that the number of cattle
Equationupper C 1
one year later would be

Equationupper C 1 equals upper C 0 plus 40 percent-sign
of
Equationupper C 0 equals 1 dot upper C 0 plus 0.40 upper C 0 equals 1.40 upper C 0
.

For instance, if there were

Equationupper C 0 equals 400
cattle initially, then
Equationupper C 1 equals 1.40 times 400 equals 560
.

But, the presence of the sheep acts to slow the rate of growth in the cattle population and, the more sheep there are, the slower the growth of the cattle population. To account for this, suppose that the population of cattle after one year is given by

Equationupper C 1 equals 1.40 upper C 0 minus 0.20 upper S 0
.

So, if there were

Equationupper C 0 equals 400
cattle and
Equationupper S 0 equals 700
sheep initially, then

Equationupper C 1 equals 1.40 times 400 minus 0.20 times 700 equals 560 minus 140 equals 420 cattle

after one year.

In a similar way, in the absence of cattle, suppose that the sheep population would grow by 50% per year, say, so that the number of sheep

Equationupper S 1
after one year would be
Equationupper S 1 equals 1.50 upper S 0
. However, the presence of cattle slows the rate of growth of the sheep. Suppose we have

Equationupper S 1 equals 1.50 upper S 0 minus 0.30 upper C 0

after one year instead.

Example 2 a. Write the pair of equations giving the populations of sheep and cattle one year later based on the populations this year as a matrix equation.

b. Assume that there are 400 cattle and 700 sheep in the valley at the start of this process. Use the matrix equation from part (a) to predict the two populations after one year.

Solution a. We write the initial numbers of cattle and sheep as a vector of populations,

Equationupper P 0 equals StartBinomialOrMatrix upper C 0 Choose upper S 0 EndBinomialOrMatrix
. Similarly, we write the populations next year also as a vector of populations,
Equationupper P 1 equals StartBinomialOrMatrix upper C 1 Choose upper S 1 EndBinomialOrMatrix
. Finally, we need a matrix A to represent the way that the populations this year are transformed into the populations one year later. We have the two equations

Equationupper C 1 equals 1.40 upper C 0 minus 0.20 upper S 0

Equationupper S 1 equals 1.50 upper S 0 minus 0.30 upper C 0 period

To set up the matrix, we must keep the order of the variables consistent in the two equations. Therefore, we rewrite the order of the terms in the second equation, so that we work with the equivalent pair of equations

Equationupper C 1 equals 1.40 upper C 0 minus 0.20 upper S 0

Equationupper S 1 equals minus 0.30 upper C 0 plus 1.50 upper S 0 period

The corresponding matrix of coefficients A, which is called the transition matrix , is then

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1.40 2nd Column negative 0.20 2nd Row 1st Column negative 0.30 2nd Column 1.50 EndMatrix period

The pair of population equations can then be written as the simple matrix equation

Equationupper P 1 equals upper A upper P 0
.

b. If the starting populations are

Equationupper C 0 equals 400
cattle and
Equationupper S 0 equals 700
sheep, then we have the starting population vector
Equationupper P 0 equals StartBinomialOrMatrix 400 Choose 700 EndBinomialOrMatrix period
Using the matrix multiplication feature of a calculator, the population vector one year later is

Equationupper P 1 equals upper A upper P 0 equals Start 2 By 2 Matrix 1st Row 1st Column 1.40 2nd Column negative 0.20 2nd Row 1st Column negative 0.30 2nd Column 1.50 EndMatrix StartBinomialOrMatrix 400 Choose 700 EndBinomialOrMatrix equals StartBinomialOrMatrix 420 Choose 930 EndBinomialOrMatrix equals StartBinomialOrMatrix upper C 1 Choose upper S 1 EndBinomialOrMatrix
.

Notice that the number of cattle, 420, is the same value that we calculated before. We therefore see that, despite the competition, the cattle population actually increases somewhat (from 400 to 420) while the sheep population increases considerably (from 700 to 930).

Later in this section we will discuss the way that matrix multiplication is defined; for now, we only cite the results based on the use of technology.

The results of the above example raise some intriguing questions. For instance, can we figure out what happens after a second year? It is actually very simple to answer this using matrix methods. The transition matrix A connects the two populations between any two successive years:

EquationPopulation vector next year equals upper A dot Population vector this year period

In particular, the population vector,

Equationupper P 2
, after a second year is related to the population vector,
Equationupper P 1
, after one year via the matrix equation
Equationupper P 2 equals upper A upper P 1
. Again doing the matrix multiplication on a calculator, we find that

Equationupper P 2 equals upper A upper P 1 equals Start 2 By 2 Matrix 1st Row 1st Column 1.40 2nd Column negative 0.20 2nd Row 1st Column negative 0.30 2nd Column 1.50 EndMatrix StartBinomialOrMatrix 420 Choose 930 EndBinomialOrMatrix equals StartBinomialOrMatrix 402 Choose 1269 EndBinomialOrMatrix
.

Thus, the cattle population has decreased slightly (from 420 to 402), while the sheep population has grown even more dramatically (from 930 to 1269).

You might wonder what happens over the course of still more years. Does the cattle population continue to decrease and eventually die out? Does the sheep population continue to increase more and more rapidly? To answer questions such as this, we can continue to apply the same matrix analysis, but we need some better notation. We have the matrix equations:

Equationupper P 1 equals upper A upper P 0

Equationupper P 2 equals upper A upper P 1

Equationupper P 3 equals upper A upper P 2

and so on. Alternatively, we can write these matrix equations in a somewhat more suggestive manner as

Equationupper P 1 equals upper A upper P 0

Equationupper P 2 equals upper A upper P 1 equals upper A left-parenthesis upper A upper P 0 right-parenthesis equals left-parenthesis upper A upper A right-parenthesis upper P 0
Substitute
Equationupper P 1

Equationupper P 3 equals upper A upper P 2 equals upper A left-parenthesis upper A upper A right-parenthesis upper P 0 equals left-parenthesis upper A upper A upper A right-parenthesis upper P 0
Substitute
Equationupper P 2

and so forth. This suggests that we might be able to write these equations using powers of the matrix A with

Equationupper A squared equals upper A upper A
,
Equationupper A cubed equals upper A upper A upper A
, and so on. We will discuss how powers of matrices are defined later in this section; for now, we simply show how they are used. Thus, in the matrix equation
Equationupper P 2 equals left-parenthesis upper A upper A right-parenthesis upper P 0
, we write
Equationleft-parenthesis upper A upper A right-parenthesis equals upper A squared
. Using

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1.40 2nd Column negative 0.20 2nd Row 1st Column negative 1.30 2nd Column 1.50 EndMatrix

and the matrix capabilities of a calculator, we find that

Equationupper A squared equals Start 2 By 2 Matrix 1st Row 1st Column 2.02 2nd Column negative 0.58 2nd Row 1st Column negative 0.87 2nd Column 2.31 EndMatrix

and therefore

Equationupper P 2 equals upper A squared upper P 0 equals StartBinomialOrMatrix 402 Choose 1269 EndBinomialOrMatrix
,

which is the same set of population values we found previously for

Equationupper P 2
.

We can continue to do this for each successive year, getting

Equationupper P 3 equals upper A cubed upper P 0
,
Equationupper P 4 equals upper A Superscript 4 Baseline upper P 0
, and so on, as illustrated in the next example.

Example 3 Find the populations of cattle and sheep in this valley

a. after 3 years;

b. after 4 years;

c. after 5 years.

Interpret the results of each.

Solution a. To find the populations after 3 years, we need

Equationupper A cubed upper P 0
. Using a calculator, we get

Equationupper P 3 equals upper A cubed upper P 0 equals StartBinomialOrMatrix 309 Choose 1282.9 EndBinomialOrMatrix almost-equals StartBinomialOrMatrix 309 Choose 1283 EndBinomialOrMatrix
.

Consequently, the cattle population has decreased rather considerably during the third year (from 402 to 309) while the sheep population has increased by very little (from 1269 to 1283).

b. To find the populations after 4 years, we calculate

Equationupper A Superscript 4 Baseline upper P 0
, which is

Equationupper P 4 equals upper A Superscript 4 Baseline upper P 0 equals StartBinomialOrMatrix 76.02 Choose 2581.6 EndBinomialOrMatrix almost-equals StartBinomialOrMatrix 76 Choose 2582 EndBinomialOrMatrix
.

We therefore see that the cattle population has decreased very significantly while the sheep population has roughly doubled. Apparently, the growth in the sheep has come at the expense of the cattle.

c. To find the populations after 5 years, we calculate

Equationupper A Superscript 5 Baseline upper P 0
, which is

Equationupper P 5 equals upper A Superscript 5 Baseline upper P 0 equals StartBinomialOrMatrix negative 409.9 Choose 3849.7 EndBinomialOrMatrix
.

Clearly, it is not possible for the number of cattle to be negative. Therefore, according to this mathematical model, the cattle population will have died out sometime during the fifth year and the sheep population would be more than five times the original number of 700 sheep. This may be unreasonable because there is no obvious reason why none of the original 400 cattle, let alone their subsequent offspring, will have survived for five years; after all, there is no indication of any predators that will eliminate the cattle. So perhaps the values used for the coefficients in the original model are not particularly accurate for modeling this situation or perhaps the model itself is too simple.

Another interesting issue we can study is the effect of the initial populations on the subsequent population values. Do the cattle always die out and the sheep flourish, as in Example 3, or might other situations hold? We ask you to explore this question in some of the problems at the end of this section.

Comparing Successive Vectors

Earlier in this section, we introduced a geometric interpretation for vectors with two components—they can be viewed as line segments starting at the origin. If the vector is

Equationupper X equals StartBinomialOrMatrix x Choose y EndBinomialOrMatrix
, then the corresponding line segment extends from the origin to the point (x , y ) and has magnitude

Equationdouble-vertical-bar upper X double-vertical-bar equals StartRoot x squared plus y squared EndRoot
.

We also gave a geometric interpretation to the product

Equationupper A upper X
of a
Equation2 times 2
matrix A and a vector X with two components: the act of multiplying A and X produces the new vector
Equationupper Upsilon equals upper A upper X
, which we picture as a different line segment starting at the origin. In Example 3, we constructed a sequence of such products involving the transition matrix A and the population vectors
Equationupper P 0
,
Equationupper P 1 equals upper A upper P 0
,
Equationupper P 2 equals upper A upper P 1
,
Equationupper P 3 equals upper A upper P 2
, and so on, so that we have

Equationupper P 1 equals upper A upper P 0

Equationupper P 2 equals upper A upper P 1 equals upper A squared upper P 0
,

Equationupper P 3 equals upper A upper P 2 equals upper A cubed upper P 0
,

Equationupper P 4 equals upper A upper P 3 equals upper A Superscript 4 Baseline upper P 0
,

and so forth. How do these successive vectors compare to one another geometrically? We look at this in the next example.

Example 4 Consider the

Equation2 times 2
transition matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 0.8 2nd Column 0.2 2nd Row 1st Column 0.3 2nd Column 0.5 EndMatrix

and the initial vector

Equationupper X 0 equals StartBinomialOrMatrix 10 Choose 10 EndBinomialOrMatrix
. Find and describe how the subsequent vectors
Equationupper X 1 equals upper A upper X 0
,
Equationupper X 2 equals upper A upper X 1
,
Equationupper X 3 equals upper A upper X 2
, . . . are related to each other.

Solution We picture the initial vector

Equationupper X 0 equals StartBinomialOrMatrix 10 Choose 10 EndBinomialOrMatrix
as starting at the origin and extending to the point Equationleft-parenthesis 10 comma 10 right-parenthesis. Using a calculator, we find that

Equationupper X 1 equals upper A upper X 0 equals StartBinomialOrMatrix 10 Choose 8 EndBinomialOrMatrix
,

Equationupper X 2 equals upper A upper X 1 equals upper A squared upper X equals StartBinomialOrMatrix 9.6 Choose 7 EndBinomialOrMatrix
,

Equationupper X 3 equals upper A upper X 2 equals upper A cubed upper X equals StartBinomialOrMatrix 9.08 Choose 6.38 EndBinomialOrMatrix
,

Equationupper X 4 equals upper A upper X 3 equals upper A Superscript 4 Baseline upper X equals StartBinomialOrMatrix 8.54 Choose 5.91 EndBinomialOrMatrix
,

and so forth. We show these vectors in Figure 4.11. Notice that each successive vector is slightly shorter than the one before it. More importantly, notice that, although the inclination of each successive vector changes, there is a relatively large change in the angle between

Equationupper X 0
and
Equationupper X 1
, then a considerably smaller change in the angle between
Equationupper X 1
and
Equationupper X 2
, and a still smaller change in the angle between
Equationupper X 2
and
Equationupper X 3
. In fact, the change in angle between
Equationupper X 2
and
Equationupper X 3
, and the changes in angles between all subsequent vectors, is so small that you can't distinguish one vector from the other, so that they appear to be on top of each other.

Figure 4.11

This image shows a graph with 4 vectors labeled from X Sub 0 to X Sub 3. The graph has a y and x axis that ranges from 0 to 10. All 4 vectors start from the origin, X sub 0 ending the highest at 10, 10. Following X sub 0 is X sub 1, then X sub 2, then X sub 3.

The results of Example 4 are typical of what happens when you apply the successive powers of a matrix A to a starting vector X. Each successive power transforms the vector into a different vector, one usually having a different length and a different direction. But the directions usually become almost indistinguishable from one another and, quite quickly, the successive vectors soon appear to be one atop the other as they eventually all point in the same direction.

We next consider a similar example involving

Equation3 times 3
matrices.

Example 5 Since most households already subscribe to some form of "cable" TV service, the three providers of these services—cable, satellite, and phone—can expand only by taking customers away from one of the other services and so there is intense competition in many areas. A study in one area has found that, in any given year, of those who subscribe to cable TV service, 75% remain with cable, 10% switch to satellite, and 15% switch to phone. Also, of those who subscribe to satellite TV service, 20% switch to cable, 60% remain with satellite, and 20% switch to phone. Finally, of those who subscribe to phone TV service, 20% switch to cable, 10% switch to satellite, and 70% remain with phone. In this service area, there are currently 200,000 households who subscribe to cable service, 100,000 who subscribe to satellite service, and 150,000 who subscribe to phone service.

a. Set up the transition matrix A and the initial vector

Equationupper X 0
for a matrix model for this situation.

b. Calculate the number of households in this area who will subscribe to each service next year.

c. Calculate the number of households in this area who will subscribe to each service the following year.

d. What happens in the long run in this area?

Solution a. The given information leads to the

Equation3 times 3
transition matrix

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 0.75 2nd Column 0.20 3rd Column 0.20 2nd Row 1st Column 0.10 2nd Column 0.60 3rd Column 0.10 3rd Row 1st Column 0.15 2nd Column 0.20 3rd Column 0.70 EndMatrix
,

where we express the values as decimals rather than as percentages. The initial vector for the number of subscribers to each service, in thousands, is

Equationupper X 0 equals Start 3 By 1 Matrix 1st Row 200 2nd Row 100 3rd Row 150 EndMatrix
.

b. The vector

Equationupper X 1 equals upper A upper X 0
giving the number of households subscribing to each form of TV service after one year is

Equationupper X 1 equals upper A upper X 0 equals Start 3 By 1 Matrix 1st Row 200 2nd Row 95 3rd Row 155 EndMatrix
.

Notice that the number of cable subscribers has remained the same, but that the number of satellite subscribers has decreased and that the number of phone subscribers has increased by the same amount.

c. The vector

Equationupper X 2 equals upper A upper X 1
giving the number of households subscribing to each service after a second year is

Equationupper X 2 equals upper A upper X 1 equals Start 3 By 1 Matrix 1st Row 200 2nd Row 92.5 3rd Row 157.5 EndMatrix
.

Notice that the number of cable subscribers again has remained the same, but that the number of satellite subscribers has decreased further and that the number of phone subscribers has increased by the same amount.

d. If we apply the transition matrix A repeatedly to the successive vectors, we find that they become closer and closer to the vector

EquationStart 3 By 1 Matrix 1st Row 200 2nd Row 90 3rd Row 160 EndMatrix
.

Thus, eventually, there will be 200,000 households who subscribe to cable service, 90,000 who subscribe to satellite service, and 160,000 who subscribe to phone service.

How the Product of a Matrix and a Vector is Defined

Consider the system of two linear equations in two unknowns:

Equation3 x plus 4 y equals 10
,

Equation5 x minus 6 y equals 4
.

Using the approach in Section 4.1, we convert this system into the matrix equation

Equationupper A upper X equals upper B
with

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 4 2nd Row 1st Column 5 2nd Column negative 6 EndMatrix
,
Equationupper X equals StartBinomialOrMatrix x Choose y EndBinomialOrMatrix
, and
Equationupper B equals StartBinomialOrMatrix 10 Choose 4 EndBinomialOrMatrix
,

so that

Equationupper A upper X equals Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 4 2nd Row 1st Column 5 2nd Column negative 6 EndMatrix StartBinomialOrMatrix x Choose y EndBinomialOrMatrix equals StartBinomialOrMatrix 10 Choose 4 EndBinomialOrMatrix equals upper B

Let's look carefully at how the product of A and X is formed to produce B. First, look at the left-hand side,

Equation3 x plus 4 y
, of the first equation
Equation3 x plus 4 y equals 10
. This comes from the following steps:

1. We take the product of 3, the first entry in the first row of matrix A, with x , the first entry in the column vector X;

2. Then, we take the product of 4, the second entry in the first row of A, with y , the second entry in the column vector X;

3. Finally, we add the two terms together to get

Equation3 x plus 4 y
.

We perform the same three steps with the second row of matrix A consisting of the entries 5 and

Equationnegative 6
. That is, we multiply 5 by x and
Equationnegative 6
by y and add them to produce
Equation5 x minus 6 y
. The same process lets us multiply any n by n square matrix and any n by 1 column vector X, as we illustrate with a
Equation3 times 3
matrix in the following example.

Example 6 In Example 4 of Section 4.1, we converted the system of linear equations

Equation5 x minus 7 y plus 4 z equals 6
(1)

Equation2 x plus 4 y minus 8 z equals 13
(2)

Equation3 x minus 5 y minus 9 z equals negative 2
(3)

into the corresponding matrix equation

Equationupper A upper X equals upper B
, where

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 5 2nd Column negative 7 3rd Column 4 2nd Row 1st Column 2 2nd Column 4 3rd Column negative 8 3rd Row 1st Column 3 2nd Column negative 5 3rd Column negative 9 EndMatrix
,
Equationupper X equals Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix
, and
Equationupper B equals Start 3 By 1 Matrix 1st Row 6 2nd Row 13 3rd Row negative 2 EndMatrix
.

Therefore

Equationupper A equals Start 3 By 3 Matrix 1st Row 1st Column 5 2nd Column negative 7 3rd Column 4 2nd Row 1st Column 2 2nd Column 4 3rd Column negative 8 3rd Row 1st Column 3 2nd Column negative 5 3rd Column negative 9 EndMatrix Start 3 By 1 Matrix 1st Row x 2nd Row y 3rd Row z EndMatrix equals Start 3 By 1 Matrix 1st Row 6 2nd Row 13 3rd Row negative 2 EndMatrix equals upper B

Explain the process by which the product of A and X is formed to produce B.

Solution Consider first the left-hand side,

Equation5 x minus 7 y plus 4 z
, of Equation (1). This is produced by:

1. Multiplying 5, the first entry in the first row of A, by x , the first entry in vector X;

2. Multiplying

Equationnegative 7
, the second entry in the first row of A, by y , the second entry in vector X;

3. Multiplying 4, the third entry of the first row of A, by z , the third entry in vector X.

4. Adding the three terms together to get

Equation5 x minus 7 y plus 4 z
, which is equal to 6, the first entry in B.

We do the same four steps with the second row of matrix A consisting of the entries 2, 4, and

Equationnegative 8
. That is, we multiply 2 by x , 4 by y , and
Equationnegative 8
by z and add them to produce
Equation2 x plus 4 y minus 8 z
. This must equal 13, the second component of B. Similarly, with the third row of A consisting of 3,
Equationnegative 5
, and
Equationnegative 9
, we multiply 3 by x ,
Equationnegative 5
by y , and
Equationnegative 9
by z and add them to produce
Equation3 x minus 5 y minus 9 z
, which must equal
Equationnegative 2
, the third component of B.

How the Product of Two Matrices is Defined

Now suppose we want to multiply two square matrices A and Β of the same size to form AB in a natural way; that is, a way that is an obvious extension of what we have just done with the product of a matrix and a vector. We consider

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 4 2nd Row 1st Column 5 2nd Column negative 6 EndMatrix
and
Equationupper B equals Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 7 2nd Row 1st Column negative 1 2nd Column 9 EndMatrix
,

so that

Equationupper A upper B equals Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 4 2nd Row 1st Column 5 2nd Column negative 6 EndMatrix Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 7 2nd Row 1st Column negative 1 2nd Column 9 EndMatrix
.

Think of the second matrix B as being composed of two different column vectors:

Equationupper X 1 equals StartBinomialOrMatrix 2 Choose negative 1 EndBinomialOrMatrix
and
Equationupper X 2 equals StartBinomialOrMatrix 7 Choose 9 EndBinomialOrMatrix
.

We multiply each of these vectors by the matrix A and combine the results into a new matrix. We diagram the process in Figure 4.12. We start (#1) by multiplying the first row of A by the first column

Equationupper X 1
of matrix B to get

Equation3 times 2 plus 4 times left-parenthesis negative 1 right-parenthesis equals 6 minus 4 equals 2
.

We then (#2) multiply the second row 5 and

Equationnegative 6
of A by the first column
Equationupper X 1
of B to get

Equation5 times 2 plus left-parenthesis negative 6 right-parenthesis times left-parenthesis negative 1 right-parenthesis equals 10 plus 6 equals 16
.

This produces the first column of AB with the two entries 2 and 16.

Figure 4.12

This image shows two matrixes, the first one being Start 2 by 2 Matrix left bracket Row 3 4 Row 5 Minus 6 right bracket and the second one being Start 2 by 2 Matrix left bracket Row 2 7 Row Minus 1 9 right bracket. There are 4 numbered arrows, #1 pointing to both the 4 in the first matrix and the 2 in the second matrix. #2 pointing from Minus 6 in the first matrix to Minus 1 in the second matrix. #3 pointing to 3 in the first matrix to 7 in the second matrix and #4 points to 5 in the first matrix to 9 in the second matrix.

Having multiplied matrix A by the first column

Equationupper X 1
of matrix B, we then multiply matrix A by the second column
Equationupper X 2
of B consisting of 7 and 9 in an identical way. We therefore obtain

(#3):

Equation3 times 7 plus 4 times 9 equals 21 plus 36 equals 57
,

(#4):

Equation5 times 7 plus left-parenthesis negative 6 right-parenthesis times 9 equals 35 minus 54 equals negative 19
,

and so the second column of the product AB consists of the entries 57 and

Equationnegative 19
. That is,

Equationupper A upper B equals Start 2 By 2 Matrix 1st Row 1st Column 3 2nd Column 4 2nd Row 1st Column 5 2nd Column negative 6 EndMatrix Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 7 2nd Row 1st Column negative 1 2nd Column 9 EndMatrix equals Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 57 2nd Row 1st Column 16 2nd Column negative 19 EndMatrix
.

This is the same result you would get using your calculator to calculate

Equationupper A upper B
directly.

The technique outlined above can be extended in a variety of ways. First, it can be applied to form the product of any two square matrices of the same dimensions, either 3 by 3 or 4 by 4 or larger. Second, it can be used to form the successive powers

Equationupper A squared
,
Equationupper A cubed
,
Equationupper A Superscript 4
, . . . of any square matrix. In fact, it can also be extended, with certain limitations, to form the products of non-square matrices, although we do not discuss that here.

What is the Inverse Matrix?

We now discuss the meaning of the inverse

Equationupper A Superscript negative 1
, if it exists, for an
Equationn times n
square matrix A. To begin, when we multiply numbers in arithmetic, the number 1 plays a special role-it is the only number that has the property that, for any number a whatsoever,
Equation1 times a equals a times 1 equals a
. Mathematically, we describe the number 1 as the multiplicative identity . Moreover, for any number a other than 0, there is an associated number
Equation1 slash a
such that
Equationleft-parenthesis 1 slash a right-parenthesis times a equals a times left-parenthesis 1 slash a right-parenthesis equals 1
, and the number
Equation1 slash a
is called the multiplicative inverse of a .

When we multiply

Equation2 times 2
matrices, there is a particular matrix that plays the same role as the number 1 does in arithmetic. It is the matrix

Equationupper I equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndMatrix

and it is called the multiplicative identity matrix or simply the identity matrix . To see why, consider any other

Equation2 times 2
matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix
.

When we multiply A and I, we get

Equationupper A upper I equals Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndMatrix equals Start 2 By 2 Matrix 1st Row 1st Column a dot 1 plus b dot 0 2nd Column a dot 0 plus b dot 1 2nd Row 1st Column c dot 1 plus d dot 0 2nd Column c dot 0 plus d dot 1 EndMatrix equals Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix equals upper A
,

so that the product of any 2 by 2 matrix A and I is A.

This image shows a statue being used as a bulletpoint

Thought Experiment

Show that, for any 2 by 2 matrix A,

Equationupper I upper A equals upper A
also.

Furthermore, most (but certainly not all) square matrices A have an inverse matrix

Equationupper A Superscript negative 1
that has the property that

Equationupper A Superscript negative 1 Baseline upper A equals upper A upper A Superscript negative 1 Baseline equals upper I
.

For instance, the matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 4 2nd Row 1st Column 4 2nd Column 0 EndMatrix

has an inverse matrix

Equationupper A Superscript negative 1
that turns out to be, using a calculator,

Equationupper A Superscript negative 1 Baseline equals Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column one-fourth 2nd Row 1st Column one-fourth 2nd Column negative one-eighth EndMatrix
.

To verify that this is indeed the inverse, we multiply this matrix by A to get

Equationupper A Superscript negative 1 Baseline upper A equals Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column one-fourth 2nd Row 1st Column one-fourth 2nd Column negative one-eighth EndMatrix Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 4 2nd Row 1st Column 4 2nd Column 0 EndMatrix equals Start 2 By 2 Matrix 1st Row 1st Column 0.2 plus one-fourth dot 4 2nd Column 0 dot 4 plus one-fourth dot 0 2nd Row 1st Column one-fourth dot 2 minus one-eighth dot 4 2nd Column one-fourth dot 4 minus one-eighth dot 0 EndMatrix equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndMatrix equals upper I

Similarly, we can check that

Equationupper A upper A Superscript negative 1 Baseline equals upper I
. While we will not go into methods for finding the inverse of a matrix here, we will ask you to find this particular inverse matrix in the problem set.

As we said previously, some matrices do not have inverses. For instance, the matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column negative 4 2nd Row 1st Column negative 2 2nd Column 8 EndMatrix

does not have an inverse. If you ask a calculator or software package to calculate

Equationupper A Superscript negative 1
, it will respond with an error message such as "SINGULAR MATRIX". We ask you to investigate this matrix in Problem 16.

This image shows a statue being used as a bulletpoint

Thought Experiment

What would you expect the identity matrix for any 3 by 3 matrix to be? What property would it have? What property would the inverse

Equationupper A Superscript negative 1
of a 3 by 3 matrix have, if it exists?

Problems

In problems

Equation1 minus 5
, find the length (magnitude) of each vector.

1.

EquationStartBinomialOrMatrix 3 Choose 4 EndBinomialOrMatrix

2 .

EquationStartBinomialOrMatrix 3 Choose negative 4 EndBinomialOrMatrix

3.

EquationStartBinomialOrMatrix 5 Choose 12 EndBinomialOrMatrix

4.

EquationStartBinomialOrMatrix 2 Choose 7 EndBinomialOrMatrix

5.

EquationStart 3 By 1 Matrix 1st Row 1 2nd Row 3 3rd Row 6 EndMatrix

6. The mathematical model introduced in the text for cattle and sheep was based on an initial population of 400 cattle and 700 sheep. Suppose the two starting population values are interchanged so that there are initially 700 cattle and 400 sheep. Use the matrix equation to calculate the number of cattle and sheep in each of the following three years. Does the cattle population still decline and the sheep population still grow? If so, roughly how long does it now take for the cattle population to die out?

7. Repeat Problem 6 if the starting populations are 800 cattle and 500 sheep.

8. Repeat Problem 6 if the starting populations are 500 cattle and 500 sheep.

9. Consider the mathematical model for the population of cattle and sheep with the pair of equations

Equationupper C 1 equals 1.40 upper C 0 minus 0.25 upper S 0
,

Equationupper S 1 equals minus 0.20 upper C 0 plus 1.50 upper S 0
,

instead of the original pair of equations

Equationupper C 1 equals 1.40 upper C 0 minus 0.20 upper S 0
,

Equationupper S 1 equals minus 0.30 upper C 0 plus 1.50 upper S 0
.

a. What is the practical meaning of the change in the coefficient of Equationupper S 0 in the first equation? What do you expect might happen with the populations over the following years because of this change?

b. Calculate the population of cattle and sheep, using the same starting population values of 400 cattle and 700 sheep, (i) after 1 year; (ii) after 2 years; (iii) after 3 years.

c. If the starting populations are 600 cattle and 300 sheep, repeat the calculations for the populations (i) after 1 year; (ii) after 2 years; (iii) after 3 years using both models.

10. In Example 5, we explored the distribution of households subscribing to different forms of cable TV service. Suppose now that there are 225,000 cable subscribers, 85,000 satellite subscribers, and 140,000 phone subscribers, though the transition matrix remains the same. Find the number of subscribers to each service

a. after one year;

b. after two years.

c. Find the limiting number of subscribers to each service as time passes.

d. Does the limiting number of subscribers seem to depend on the initial distribution or on the transition matrix?

11. Suppose that, of those who subscribe to cable TV service in some area, 70% remain with cable, 10% switch to satellite, and 20% switch to phone. Also, of those who subscribe to satellite TV service, 15% switch to cable, 60% remain with satellite, and 25% switch to phone. Finally, of those who subscribe to phone TV service, 15% switch to cable, 10% switch to satellite, and 75% remain with phone. In this service area, there are currently 200,000 households who subscribe to cable service, 100,000 who subscribe to satellite service, and 150,000 who subscribe to phone service. Find the number of subscribers to each service

a. after one year;

b. after two years.

c. Find the limiting number of subscribers to each service as time passes.

d. Does the limiting number of subscribers seem to depend on the initial distribution or on the transition matrix?

12. Repeat Problem 11 if the initial distribution of households is the same as in Problem 10.

13. The matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0.2 2nd Column 0.2 EndMatrix
is applied successively to the vector
Equationupper X equals StartBinomialOrMatrix 2 Choose 7 EndBinomialOrMatrix
to form
Equationupper A upper X
,
Equationupper A squared upper X
and
Equationupper A cubed upper X
. Calculate and plot all three of these vectors on the same set of axes.

14. The matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column negative 2 2nd Column 3 EndMatrix
is applied successively to the vector
Equationupper X equals StartBinomialOrMatrix 2 Choose 7 EndBinomialOrMatrix
to form
Equationupper A upper X
,
Equationupper A squared upper X
and
Equationupper A cubed upper X
. Calculate and plot all three of these vectors on the same set of axes.

15. In this problem, you will find the inverse matrix for the matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 2 2nd Column 4 2nd Row 1st Column 4 2nd Column 0 EndMatrix
. To do this, assume that the inverse has the form
Equationupper A Superscript negative 1 Baseline equals Start 2 By 2 Matrix 1st Row 1st Column a 2nd Column b 2nd Row 1st Column c 2nd Column d EndMatrix
, where a , b , c , and d are four numbers to be determined.
Equationupper A Superscript negative 1
is the inverse matrix if it satisfies the matrix equation
Equationupper A upper A Superscript negative 1 Baseline equals upper I
. Multiply out this equation to get a system of four equations in the four unknowns (it is actually a far simpler system than you likely expect) and solve for the four unknowns. (Note: If the system of equations did not have a solution or had multiple solutions, then the matrix A would not have an inverse.)

16. In the text, we stated that the matrix

Equationupper A equals Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column negative 4 2nd Row 1st Column negative 2 2nd Column 8 EndMatrix
does not have an inverse.

a. Use a calculator to verify that this matrix does not have an inverse.

b. Attempt to find the inverse matrix to A using the same approach as in Problem 15. Describe what goes wrong so that the resulting system of linear equations does not have a solution.

4.4 Linear Models with Several Variables

Everything we have discussed up to this point has involved situations in which one variable (say y ) depends on another (say x ) in a linear fashion. In the real world things aren't always this simple and we often encounter situations in which one quantity depends on two or more other quantities. For example, when the weather report gives the wind-chill factor during the winter, that value depends on both the air temperature and the wind speed so it is a function of two independent variables. Similarly, when you take out a car loan, the monthly payment depends on the amount borrowed, the interest rate, and the length of the loan, so there are three independent variables. A study conducted at a college found that student performance in math classes could be accurately predicted from a combination of a dozen independent variables, including the student's score on a placement test, the student's age, gender, SAT score, high school GPA, number of years since the previous math course, and the grade in that course. A model based on all of them produced the most accurate predictions.

To keep things simple, suppose that we have a single variable y that depends on two independent variables

Equationx 1
and
Equationx 2
. Therefore, for each value of the first variable
Equationx 1
and each value of the second variable
Equationx 2
, there is one corresponding value for the dependent variable y . For instance, if the function is
Equationy equals left-parenthesis x 1 right-parenthesis squared plus left-parenthesis x 2 right-parenthesis Superscript 4
, then corresponding to
Equationx 1 equals 5
and
Equationx 2 equals 3
, we have
Equationy equals 5 squared plus 3 Superscript 4 Baseline equals 25 plus 81 equals 106
. We sometimes write such a function in the form
Equationy equals f left-parenthesis x 1 comma x 2 right-parenthesis
to indicate that y is a function of the two variables
Equationx 1
and
Equationx 2
. (We can easily extend all of these ideas to situations where there are three independent variables,
Equationx 1
,
Equationx 2
, and
Equationx 3
or, for that matter, n independent variables,
Equationx 1 comma x 2 comma ellipsis comma x Subscript n Baseline
, where n is any positive integer.)

Serum Cholesterol y

Weight

Equationx 1

Systolic Blood Pressure

Equationx 2

152.2

59.0

108

158.0

52.3

111

157.0

56.0

115

155.0

53.5

116

156.0

58.7

117

159.4

60.1

120

169.1

59.0

124

181.0

62.4

127

174.9

65.7

122

180.2

63.2

131

174.0

64.2

125

For instance, suppose that we want to determine whether a person's serum cholesterol level depends on his or her weight and systolic blood pressure and if it does, what the functional relationship is. (In a blood pressure measurement, such as 120 over 80, the systolic reading is the first, or higher, number, 120; the smaller number, 80, is the diastolic reading.) We collect a set of data on a random sample of individuals from some population group—say, young males. The data in the accompanying table, for a sample of 11 apparently normal males between the ages of 13 and 16, shows the weight of each of these individuals in kilograms (independent variable

Equationx 1
) and the systolic blood pressure of each (independent variable
Equationx 2
), as well as their serum cholesterol level in mg/100 cc (the dependent variable y ). If there is a relationship between the serum cholesterol level and the other two variables, we can then use it to predict the serum cholesterol level for other members of this population group, based on weight and blood pressure.

To understand the ideas involved in finding a linear function that fits this set of data and that can then be used to answer predictive questions, we have to extend some of the ideas from the previous chapters. First, if we have any ordered pair of numbers x and y , we visualize them as representing the coordinates Equationleft-parenthesis x comma y right-parenthesis of a point in the plane. Similarly, suppose we have an ordered triple of numbers, say x , y , and z , where we think of z as depending on the values of x and y . Alternatively, to allow us to extend these ideas to situations with more than two independent variables, we can write the variables as

Equationx 1
,
Equationx 2
, and y , where we think of y as depending on the values of
Equationx 1
and
Equationx 2
. We can think of these three numbers as the coordinates
Equationleft-parenthesis x 1 comma x 2 comma y right-parenthesis
of a point Ρ in three dimensional space, as shown in Figure 4.13, where there are three mutually perpendicular axes that intersect at an origin where
Equationx 1 equals x 2 equals y equals 0
. That is, the set of data is equivalent to a set of points in three-dimensional space instead of a set of points in the two-dimensional
Equationx minus y
plane.

Figure 4.13

This image shows two rectangular boxes with one corner being at the origin. Along the edges of the origin is X sub 1 and X sub 2 showing the horizontal and vertical axes respectively. The y axis goes vertically from the origin to determine the opposite corner. From the origin there is a line that points to point P which is at the top height of the box on the y axis.

Second, recall how, in Section 3.4, we fit a line to a set of Equationleft-parenthesis x comma y right-parenthesis data by finding the linear function

Equationy equals a x plus b
that comes closest to all the points in the sense that the sum of the squares of the vertical distances between the points and the line is a minimum. See Figure 4.14. Now, our data consists of points in space having coordinates
Equationleft-parenthesis x 1 comma x 2 comma y right-parenthesis
. Instead of fitting a line to a set of data in the plane, we now need to fit the three-dimensional equivalent of a line to this set of points in space. The three-dimensional object that is equivalent to a line is a plane —a flat surface with no bends or curves, like the walls or floors in a room. The equation of a plane in space, or equivalently the equation of a linear function of two independent variables
Equationx 1
and
Equationx 2
, can be written as

Figure 4.14

This image shows a diagonal line that has numberous dotted lines, 4 on top and 3 on the bottom, emerging from the base line. Each dotted line has a point at the end of it.

Equationy equals a x 1 plus b x 2 plus c
,

where a , b , and c are three constants. Similarly, if the dependent variable is z and the two independent variables are x and y , the equation of a plane is

Equationz equals a x plus b y plus c
. Both of these formulas for the equation of a plane are natural extensions of the equation
Equationy equals a x plus b
of a line in the
Equationx minus y
plane. For instance,
Equationy equals 5 x 1 plus 8 x 2 minus 11
and
Equationz equals 2 x minus 4 y plus 16
are both equations of planes in three-dimensional space. In each case, the independent variables are only multiplied by constants.

Figure 4.15.

This image shows a similar figure as Figure 4.14. However, the base line is now a Rhombus, with 4 lines emerging from the top and 3 emerging from the bottom of the Rhombus. Each new line contains a point at the end of them.

Our objective is to find the linear function that is the best fit to the set of three-dimensional data points. That is, we want to find the plane in space that comes closest to all of the

Equationleft-parenthesis x 1 comma x 2 comma y right-parenthesis
data points, as shown in Figure 4.15. This best-fit plane is called the regression plane . The process of finding the equation of this plane is called multivariate regression or multiple regression . If there are more than two independent variables, a natural extension is used although it isn't possible to visualize the "hyperplane" in four or more dimensions that best fits these data points.

The calculations involved in any multivariate regression are extremely tedious, but the method is so widely used that a routine is available in almost any statistical software package and in most spreadsheets and on a few hand-held calculators. We will not be concerned with the mechanics of calculating these quantities, but will simply cite the results obtained by using appropriate software. In the problems at the end of this section, we similarly assume that you have access to software that performs the calculations for you. Later in the section, we outline how to use Excel to perform such calculations.

Example 1 For the data on serum cholesterol level of young males versus their body weights

Equationleft-parenthesis x 1 right-parenthesis
and systolic blood pressures
Equationleft-parenthesis x 2 right-parenthesis
,

a. Find the multivariate regression equation expressing serum cholesterol level as a function of both body weight and systolic blood pressure.

b. Interpret the coefficients in the multivariate regression equation.

c. Use the regression equation to predict the serum cholesterol level of a male in the 13- to 16-year-old age group who weighs 60 kg and whose systolic blood pressure is 123.

Solution a. Using an appropriate software package, we find that the multivariate regression equation is

Equationy equals negative 7.6419 plus 0.6297 x 1 plus 1.1315 x 2
,

or equivalently

Cholesterol Level

Equationequals negative 7.6419 plus 0.6297 times weight plus 1.1315 times
systolic blood pressure or, if we choose to use letters that reflect the quantities involved

Equationupper C equals negative 7.6419 plus 0.6297 upper W plus 1.1315 upper P period

b. To interpret this regression equation, let's see what happens to the values for the cholesterol level when we change one of the independent variables, either the weight or the systolic blood pressure. In particular, suppose that a person's weight W or

Equationx 1
increases by 1 kg. According to the formula, the cholesterol level C will increase by 0.6297 units because the coefficient of W is 0.6297. Alternatively, suppose that a person's systolic blood pressure P or
Equationx 2
increases by 1 unit. In that case, the cholesterol level C will increase by 1.1315 units. Thus an increase of 1 unit in blood pressure has a considerably larger impact on cholesterol level than an increase of 1 unit in weight has on cholesterol level. That is, for a 1-unit increase in each quantity, the systolic blood pressure apparently has a greater effect on cholesterol level than weight does.

c. According to this multivariate linear regression model, we predict that the serum cholesterol level for an individual who weighs

Equationupper W equals 60 kg
and whose systolic blood pressure
Equationupper Rho equals 123
is

Equationupper C equals negative 7.6419 plus 0.6297 times 60 plus 1.1315 times 123 equals 169.315

or about 169 mg per Equation100 c c.

Note how the coefficients

Equationa equals 0.6297
and
Equationb equals 1.1315
play the same role in this multivariate regression equation
Equationy equals a x 1 plus b x 2 plus c
that the slope m plays in the equation of a line
Equationy equals m x plus b
. We can think of the regression plane as having two different slopes, one in the
Equationx 1
direction and another in the
Equationx 2
direction, and they indicate how quickly the dependent variable increases or decreases for a given change in either
Equationx 1
or
Equationx 2
, respectively. Imagine that you are standing on the side of a hill. Suppose that the
Equationx 1
direction is to the east and the
Equationx 2
direction is to the north. If you take a step eastward in the
Equationx 1
direction, there is a certain rise and so you can measure the slope in that direction and get a value, call it
Equationm 1
. Alternatively, if you take the same size step northward in the
Equationx 2
direction, there is a different rise (or perhaps a drop) and so you can measure the slope in that direction and get a different value, call it
Equationm 2
. Thus, there are different slopes, one with respect to the
Equationx 1
-axis and the other with respect to the
Equationx 2
-axis. For that reason, you may want to think of the equation of the regression plane in the form

Equationy equals m 1 x 1 plus m 2 x 2 plus c
,

where

Equationm 1
and
Equationm 2
are the slopes in the
Equationx 1
and
Equationx 2
directions, respectively. For instance, if you are standing at a height of 50 feet on the side of the hill and the slope to the east (
Equationx 1
direction) is 2 and the slope to the north (
Equationx 2
direction) is 1, the equation of the plane would be
Equationy equals 2 x 1 plus 1 x 2 plus 50
.

Also note that, in making the prediction in part (c) of Example 1, we took values for the independent variables

Equationx 1
and
Equationx 2
that were within the ranges of the data set. Had we used values that were well outside the ranges of the data (say,
Equationx 1 equals 70 kg
and
Equationx Subscript 2 Baseline equals 140
), then we would have considerably less confidence in the accuracy of the prediction.

The Multiple Correlation Coefficient

In addition to finding the equation of the plane that is the best fit to a set of data points in three-dimensional space, we also need a way of measuring how good the fit is. With Equationleft-parenthesis x comma y right-parenthesis data points (where y is a function of x ), we used the linear correlation coefficient r as such a measure. With two or more independent variables, we use a comparable quantity known as the multiple correlation coefficient , denoted by R . Like r , R takes on values between

Equationnegative 1
and 1, and the closer R is to either
Equationplus 1
or
Equationnegative 1
, the stronger the linear relationship between y and the linear combination
Equationa x 1 plus b x 2 plus c
of the independent variables.

Furthermore, the square of the multiple correlation coefficient,

Equationupper R squared
, is known as the coefficient of determination . The value of
Equationupper R squared
provides extremely useful information about the extent to which the multivariate regression equation explains the relationship between the dependent variable y and the independent variables
Equationx 1
and
Equationx 2
(if there are two) or
Equationx 1 comma x 2 comma ellipsis comma x Subscript n Baseline
, (if there are
Equationn greater-than 2
). In particular, the value of
Equationupper R squared
indicates the percentage of the variation in the data that is explained by the linear relationship. For instance, if
Equationupper R squared equals 0.90
, say, then 90% of the variation is explained by the linear function. Consequently, 10% of the variation—the distances from the points to the plane—is due to other factors. A similar interpretation applies to
Equationr squared
, the square of the correlation coefficient with Equationleft-parenthesis x comma y right-parenthesis data points.

As with the multivariate regression equation, both the multiple correlation coefficient and the coefficient of determination are typically calculated as part of the output of spreadsheets and statistical software.

Example 2 Find and interpret both the multiple correlation coefficient and the coefficient of determination for the data in Example 1 relating serum cholesterol level to an individual's weight and systolic blood pressure.

Solution Using an appropriate software package, we find that the coefficient of determination is

Equationupper R squared equals 0.8508

and so the multiple correlation coefficient is

Equationupper R equals StartRoot upper R squared EndRoot equals StartRoot 0.8508 EndRoot equals 0.9224 period

The multiple correlation coefficient is reasonably close to 1, so we conclude that a high degree of linear correlation exists between the dependent variable y and the two independent variables

Equationx 1
(a person's weight) and
Equationx 2
(his systolic blood pressure). Statisticians have developed a table of critical values for R comparable to the critical values for r , but we won't go into that here. Moreover, the value
Equationupper R squared equals 0.8508
indicates that about 85% of the variation in serum cholesterol levels in this population group can be "explained" by these two variables.

In Example 2 the value for the coefficient of determination

Equationupper R squared equals 0.8508
indicates that about 85% of the variation observed is explained by the two variables, weight and systolic blood pressure. Thus, about 15% of the variation is not explained by these two variables. We can explain a greater amount of the variation by introducing an additional variable, perhaps the number of hours that each individual in the study spends exercising each week, since we might expect that cholesterol levels also depend on the level of physical activity. In that case, we would be working with three independent variables
Equationx 1
,
Equationx 2
, and
Equationx 1
and would obtain a linear regression equation
Equationy equals a x 1 plus b x 2 plus c x 3 plus d
that relates serum cholesterol level to all three variables. Incidentally, when we do so, the coefficients of the original two independent variables will almost certainly change.

Each time we introduce an additional variable that matters, we get a value for the coefficient of determination that is closer to

Equationplus 1
or to
Equationnegative 1
, although the change might be minimal. At the same time, we also increase the complexity of the model, and there are often drawbacks to doing so.

Performing Multivariate Regression in Excel

We now briefly introduce the use of Excel for performing a multivariate regression analysis. In Excel the dependent variable is always denoted by Y , and the different independent variables are always denoted by X . Think of the two independent variables as

Equationx 1
and
Equationx 2
. You begin by entering the given data values in the first three columns labeled A, B, and C, as shown in Figure 4.16.

Once you have entered all the data values, click Tools on the top line and scroll down to the last entry, Data Analysis . . . (We indicate the computer displays in a different font, for emphasis.) If Data Analysis . . . doesn't appear, you will have to install the Excel Analysis ToolPak™ before proceeding. To install it, click on Tools and then select Add-ins . If Analysis ToolPak is listed, just click it to permanently install it. If Analysis ToolPak isn't listed in the Add-ins dialog box, click Browse and locate the drive and folder names and the file name Analys32.xll for the Analysis ToolPak—it usually is located in the Library\Analysis folder. When you click Data Analysis . . . , you will see a long list of available statistical procedures in alphabetical order. Scroll down until you reach Regression and then either double click it or single click and then click OK. Doing so brings up the window shown in Figure 4.17.

Figure 4.16

This image shows a screenshot of Microsoft Excel - Book 2 with values entered into it. All the data is shaded except for A1, which is 152.2.

In this window, you first enter the Input Y range—the cells in which the values of the dependent variable y have been entered. The simplest way to do this is to click the icon at the right end of the box; this brings you back to the original spreadsheet, and you can highlight the entries down the first column under A (the y values) and then press Enter. The first box should then read: $A$1:$A$11. You then enter the Input X range—the cells in which the values of the two (or more) independent variables have been entered. Again, click the icon at the right end of the box and then highlight the entries in the second and third columns under B and C and press Enter. The second box should then show: $B$1:$C$11.

Figure 4.17

This is a screenshot of the Regression window. It shows that under Output Options the following options are Output Range: New Worksheet by Ply: and New Workbook.

Finally, you have to enter the Output range —where the results of the regression calculations will appear on the spreadsheet. You don't want them printed over the data values in the first three columns, so you probably want them printed starting, say, in column E. Click the white circle to the left of Output range and then click the icon at the right end of the box. Designate a block of cells starting at the top of column E and extending down and to the right by highlighting the first cell under E; then press Enter. Finally, in the Regression window, press OK.

Excel will then display a large amount of information, much of which is shown in Figure 4.18. Only a few of these entries are of interest to us—the ones that have been highlighted in the figure; the rest are used for more sophisticated statistical analysis. In particular, note that the first block of output is called Regression Statistics and that the first two numbers under it are labeled Multiple R and R Square —these are the values for the multiple correlation coefficient R and the coefficient of determination,

Equationupper R squared
.

The third block of output starts with three lines labeled Intercept, X Variable 1 , and X Variable 2 . The entries to their right give the vertical intercept and the coefficients of Equationx 1 and Equationx 2, respectively. In particular, if the regression equation is

Equationy equals a x 1 plus b x 2 plus c
, the Intercept is the constant coefficient c , the value corresponding to the first X Variable is the coefficient a for Equationx 1 and the value corresponding to the second X Variable is the coefficient b for Equationx 2. Once you have these values, you can write the multivariate regression equation.

Figure 4.18

This image shows a screenshot of Microsoft Excel with data already inputted. It shows the Regression Statistics which holds the Multiple R, R Square, Adjusted R Square, Standard Error and Observations and a table labeled Anova which holds the Regression, Residual, Total, Intercept, X Variable 1 and X Variable 2, all in the Summary Output.

A Table vs a List The kind of table needed to perform multivariate regression is very specific in that the data values for the dependent variable and the two or more independent variables must be listed in columns or in rows, so the table is actually a list. However, we often see tabular displays for a function of two variables that are not lists and so multivariate regression cannot be applied directly. For instance, consider the following table giving values for the wind-chill factor (you will be asked to work with this in one of the Problems at the end of the Section). For each wind speed value (

Equationupper W equals 5
, 10, 15, Equationellipsis in mph) and each temperature reading (
Equationupper T equals 35
, 40, 45, Equationellipsis in
Equationdegree upper F
), there is a corresponding value for the wind-chill factor, but these values are not arranged in a list and so you cannot perform multivariate regression on them as presented. Instead the values (or perhaps a subset of them) must be organized first into list form and then you can do the analysis. We illustrate this with a smaller table of values in Example 3 below.

Temperature

Wind speed

35

30

25

20

15

10

5

0

5

33

27

21

16

12

7

0

Equationnegative 5

10

22

16

10

3

Equationnegative 3

Equationnegative 9

Equationnegative 15

Equationnegative 22

15

16

9

2

Equationnegative 5

Equationnegative 11

Equationnegative 18

Equationnegative 25

Equationnegative 31

20

12

4

Equationnegative 3

Equationnegative 10

Equationnegative 17

Equationnegative 24

Equationnegative 31

Equationnegative 39

25

8

1

Equationnegative 7

Equationnegative 15

Equationnegative 22

Equationnegative 29

Equationnegative 36

Equationnegative 44

Example 3 Rewrite the data in the accompanying table into list format so that you can apply multivariate regression.

x

y

1

2

3

4

11

22

33

5

44

55

66

6

77

88

99

Solution For each value of

Equationx equals 1
, 2, or 3 and each value of
Equationy equals 4
, 5, or 6, there is an associated value of z . In order to apply multivariate regression, we have to recast this information in a list, either vertically or horizontally. We therefore get

x

1

1

1

2

2

2

3

3

3

y

4

5

6

4

5

6

4

5

6

Equationz equals f left-parenthesis x comma y right-parenthesis

11

44

77

22

55

88

33

66

99

In this form, you can now use Excel to do the regression analysis.

Problems

1. A study was conducted relating the adult heights of women y to the heights

Equationx 1
of their mothers and the heights
Equationx 2
of their fathers. A sample of the data from the study is shown in the accompanying table. (In the study, the measurements of the daughters' heights are to the tenth of an inch, but the heights of their parents were reported in inches.)

y

Equationx 1

Equationx 2

58.6

63

64

64.7

67

65

65.3

64

67

61.0

60

72

65.4

65

72

67.4

67

72

60.9

59

67

63.1

60

71

60.0

58

66

71.1

72

75

62.2

63

69

67.2

67

70

63.4

62

69

68.4

69

72

62.2

63

66

64.7

64

76

59.6

63

69

61.0

64

68

64.0

60

66

65.4

65

68

Source: Mario Triola, Elementary Statistics ,

Equation11 Superscript th
ed.,

Boston: Addison-Wesley, 2010.

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient based on this sample. How much of the variation in heights of the daughters in the sample is explained by the heights of their parents?

b. Find the equation of the plane that best fits the data.

c. Use your equation from part (b) to predict the height of a daughter whose mother is 62 inches tall and whose father is 71 inches tall.

2. A study was conducted relating the heights of teenage boys y to the length x of their radius bones in the forearm and the length

Equationx 2
of their femur bones (the thighbone), as shown in the accompanying table. (All measurements are in centimeters.)

y

Equationx 1

Equationx 2

149.0

21.00

42.50

152.0

21.79

43.70

155.7

22.40

44.75

159.0

23.00

46.00

163.3

23.70

47.00

166.0

24.30

47.90

169.0

24.92

48.95

174.5

25.80

50.30

176.1

26.01

50.90

176.5

26.15

50.85

179.0

26.30

51.10

Source: Wayne W. Daniel, Biostatistics: A Foundation

for A nalysis in the Health Sciences , 4th ed., New York:

John Wiley & Sons, 1987.

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in height is explained by the length of the two bones?

b. Find the equation of the plane that best fits the data.

c. Use your equation from part (b) to predict the height of a teenage boy whose radius measures 25.50 cm and whose femur measures 49.90 cm. How close does your prediction come to the boy's actual height of 172 cm?

3. A study was conducted on the Old Faithful Geyser in Yellowstone National Park to find the relationship, if any, between the height y of the geyser (in feet) and the duration

Equationx 1
of an eruption (in seconds) and the time
Equationx 2
since the previous eruption (in minutes) The data are shown in the accompanying table.

Duration

Interval

Height

Duration

Interval

Height

240

86

140

218

78

140

237

86

154

226

91

135

122

62

140

250

89

141

267

104

140

245

79

140

113

62

160

120

57

139

258

95

140

267

100

110

232

79

150

103

62

140

105

62

150

270

87

135

276

94

160

241

70

140

248

79

155

239

88

135

243

86

125

233

82

140

241

85

136

238

83

139

214

86

140

102

56

100

114

58

155

271

81

105

Source: National Park Service

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in the height of an eruption is explained by the other variables?

b. Find the equation of the plane that best fits the data.

c. Which variable, the duration or the interval since the previous eruption, has a greater effect on the height of an eruption?

d. Predict the height of an eruption of Old Faithful if it lasts 200 seconds and it has been 90 minutes since the previous eruption.

e. Suppose you arrive at Old Faithful just before an eruption that lasts for 225 seconds and reaches a maximum height of 150 feet. How long has it been since the previous eruption?

4. The temperature-humidity index (THI), or simply the heat-index, combines the temperature Τ and the relative humidity Η into a single number that shows the apparent temperature I as it feels to the body. The following chart contains a selection of different combinations of the temperature and the relative humidity and the associated value for the heat index. These values are selected from a considerably more extensive table of values that can be found in any almanac or on the web.

Temp

70

70

75

75

80

80

80

85

85

85

90

90

90

95

95

95

100

100

Hum

40

80

40

80

50

70

90

50

70

90

50

70

90

50

70

90

50

70

THI

68

71

74

78

81

84

88

88

93

102

96

106

122

107

124

150

107

144

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in the heat-index is explained by the two independent variables?

b. Find the equation of the plane that best fits the data and so approximates the relationship between the heat-index and the temperature and the humidity. (Note: The actual formula used to calculate the heat-index is not a linear function.)

c. Use your equation from part (b) to predict the heat-index corresponding to a temperature of

Equation88 degree upper F
and a relative humidity level of 83%.

d. Suppose that the air temperature is

Equation92 degree upper F
and the heat-index is 100. Use the equation of the plane to predict the relative humidity.

e. According to this model, which variable, temperature or humidity, has a greater effect on the heat-index? Explain.

5. The wind-chill factor is an adjustment made to temperature readings to take into account the effects of the wind and so indicate how cold it feels. The following table gives the wind-chill factors associated with different combinations of air temperature in degrees Fahrenheit and wind speeds in miles per hour.

Temperature

Wind speed

35

30

25

20

15

10

5

0

5

33

27

21

16

12

7

0

Equationnegative 5

10

22

16

10

3

Equationnegative 3

Equationnegative 9

Equationnegative 15

Equationnegative 22

15

16

9

2

Equationnegative 5

Equationnegative 11

Equationnegative 18

Equationnegative 25

Equationnegative 31

20

12

4

Equationnegative 3

Equationnegative 10

Equationnegative 17

Equationnegative 24

Equationnegative 31

Equationnegative 39

25

8

1

Equationnegative 7

Equationnegative 15

Equationnegative 22

Equationnegative 29

Equationnegative 36

Equationnegative 44

a. Select a variety of representative values (say a dozen) from this table to create a list (as we did in Example 3) having the same format as that in Problem 4 on the heat-index.

b. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in the wind-chill factor is explained by the two independent variables?

c. Find the equation of the plane that best fits the data and so approximates the relationship between the wind-chill factor and the temperature and the humidity. (Note: The actual formula used to calculate the wind-chill factor is not a linear function.)

d. Use your equation from part (c) to predict the wind-chill factor corresponding to a temperature of

Equation22 degree upper F
and an 18 mph wind.

e. Suppose that the air temperature is

Equation11 degree upper F
and the wind-chill factor is
Equationnegative 20
. Use the equation of the plane to predict the wind speed.

f. According to this model, which variable, temperature or wind speed has a greater effect on the wind-chill factor? Explain.

6. The body-mass index (BMI) is a measure of the fat in the human body that is based on a person's weight and height. It is widely used by physicians to assess the risk of diseases and disabilities associated with an unhealthy weight. The following table shows the BMI for various combinations of weights (in pounds) and heights (in inches).

Weight

Height

100

120

140

160

180

200

220

240

260

60

20

23

27

31

35

39

43

47

51

63

18

21

25

28

32

35

39

43

46

66

16

19

23

26

29

32

36

39

42

69

15

18

21

24

27

30

32

35

38

72

14

16

19

22

24

27

30

33

35

75

12

15

17

20

22

25

27

30

32

a. Select a variety of the values from this table and create a list with the same format as in Problem 4.

b. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in the BMI value is explained by the two independent variables?

c. Find the equation of the plane that best fits the data. (Note: The actual relationship between the variables is not a linear function.)

d. Use your equation from part (c) to predict the BMI corresponding to a person who weighs 175 pounds and who is 71 inches tall; for someone who weighs 205 pounds and is 71 inches tall.

e. According to this model, which variable, weight or height, has a greater effect on the BMI? Explain.

7. It has been found that the height H of waves on the open ocean depends on both the wind speed W and the duration D that the wind has been blowing at that speed. The following table shows the wave heights (in feet) for various combinations of wind speed (in miles per hour) and the duration of the wind (in hours).

Wind Speed W

Duration D

11.5

17.3

23.0

34.5

46.0

57.5

10

2

4

7

13

21

29

20

2

5

8

17

28

40

30

2

5

9

18

31

45

40

2

5

9

19

33

48

Source: Navarra, Atmosphere, Weather and Climate

a. Select a variety of the values from this table and create a table with the same format as in Problem 4.

b. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in the height of ocean waves is explained by the two independent variables?

c. Find the equation of the plane that best fits the data. (Note: The actual relationship between the variables is not a linear function.)

d. Use your equation from part (c) to predict the height of waves corresponding to a wind speed of 20 mph that has been blowing for 12 hours.

e. Suppose that the wind speed is 30 mph and the wave heights are 10 feet. Use the equation of the plane to predict how long the wind has been blowing at that speed.

f. According to this model, which variable, wind speed or duration, has a greater effect on the height of the waves? Explain.

8. The taste of cheddar cheese is related to the concentration of a number of chemicals, including acetic acid, hydrogen sulfide

Equationleft-parenthesis upper H Subscript 2 Baseline upper S right-parenthesis
, and lactic acid, in the cheese. A taste test was conducted on a variety of samples containing different concentrations of these three chemicals and each taster rated the cheese for "tastefulness". Some of the results are shown in the accompanying table.

Taste

Acetic Acid

Equationupper H Subscript 2 Baseline upper S

Lactic Acid

12.3

4.54

3.14

0.86

20.9

5.16

5.04

1.53

39.0

5.37

5.44

1.57

47.9

5.76

7.50

1.81

5.6

4.66

3.81

0.99

37.3

5.89

8.73

1.29

21.9

6.08

7.97

1.78

25.9

5.70

7.60

1.09

18.1

4.99

3.85

1.29

34.9

5.74

6.14

1.68

Source: Moore, David S. And George P. McCabe: Introduction to the Practice of Statistics , Freeman, 1989.

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in taste is explained by the three independent variables?

b. Find the multivariate regression equation of the hyperplane that best fits the data.

c. Use your equation from part (b) to predict the taste rating of a sample of cheddar cheese whose concentration levels of acetic acid, hydrogen sulfide, and lactic acid are 5, 6, and 1.5, respectively.

d. According to this model, which variable, acetic acid, hydrogen sulfide, or lactic acid, has the greatest effect on taste? Explain.

9. A study was conducted to find the relationship, if any, between the systolic blood pressure reading of middle-aged men and their age

Equationleft-parenthesis x 1 right-parenthesis
, their weight
Equationleft-parenthesis x 2 right-parenthesis
and the number of hours per month that they spend in meditation
Equationleft-parenthesis x 3 right-parenthesis
. The data are shown in the accompanying table.

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in systolic blood pressure is explained by the three variables?

b. Find the equation of the plane that best fits the data.

c. Use your equation from part (b) to predict the systolic blood pressure reading of a 54 year old man who weighs 200 pounds and who spends 8 hours a month in meditation.

Systolic

Equationx 1

Equationx 2

Equationx 3

141

46

207

6

153

47

215

1

137

36

190

2

139

46

210

1

135

44

214

5

139

45

226

9

133

45

237

9

150

56

229

7

131

42

179

1

146

53

217

10

10. The value of a diamond is measured by four characteristics, its carat weight, its color, its clarity, and its cut. (These are sometimes referred to as the 4C's of a diamond.) The color of a diamond is usually graded on a scale in which the letters D, E, and F indicate exceptional white coloring and subsequent letters G, H, Equationellipsis indicate various degrees of yellowish tinting. The accompanying table shows the prices of a group of diamonds and their carat weight and color, where we have given a numerical value for each color letter, so that D is listed as 1, Ε as 2, and so forth.

Carats W

Color C

Price Ρ

Carats W

Color C

Price Ρ

0.30

1

Equationdollar-sign 1302

0.50

2

Equationdollar-sign 3501

0.35

2

Equationdollar-sign 1738

0.60

6

Equationdollar-sign 4291

0.40

3

Equationdollar-sign 1911

0.70

6

Equationdollar-sign 5122

0.50

2

Equationdollar-sign 3501

0.75

1

Equationdollar-sign 7368

0.45

7

Equationdollar-sign 1 572

1.00

2

Equationdollar-sign 10588

0.60

3

Equationdollar-sign 4291

1.00

5

Equationdollar-sign 9619

0.70

3

Equationdollar-sign 5510

1.01

1

Equationdollar-sign 16008

0.80

7

Equationdollar-sign 5441

0.20

1

Equationdollar-sign 880

0.90

7

Equationdollar-sign 6682

Source: Student Project

a. Find the coefficient of determination and the multiple regression coefficient. Is there a significant level of correlation?

b. Find the equation of the regression plane that fits this data on diamond price Ρ asa function of carat weight W and color C .

c. How much of the variation in the price of a diamond can be explained by the two variables carat weight and color? What does that suggest about the relative importance of clarity and cut?

d. What is the practical significance of the fact that the coefficient of C is negative?

e. Predict the price of a diamond that weighs 0.65 carats and whose color is rated as 5 (for H).

f. Which variable, carat weight or color, has the larger impact on the price of a diamond?

11. A study was conducted to find the relationship, if any, between the amount of tar in cigarettes and the amount of nicotine they contain and the amount of carbon monoxide (CO) they have. All measurements are in milligrams per cigarette and do not include menthol or light types. The data are shown in the accompanying table.

Brand

Tar

Nicotine

Equationupper C upper O

Benson & Hedges

16

1.2

15

Camel

16

1.0

17

Capri

9

0.8

6

Carlton

1

0.1

1

Kent

13

1.0

13

Lucky Strike

13

1.1

13

Marlboro

16

1.2

15

Merit

9

0.7

11

Newport

11

0.9

15

Old Gold

18

1.4

18

Pall Mall

15

1.2

15

Raleigh

15

1.0

16

Tareyton

14

1.0

17

Viceroy

18

1.4

15

Winston

16

1.1

18

Source: Federal Trade Commission

a. Use an appropriate software package to calculate the coefficient of determination and the multiple correlation coefficient. How much of the variation in tar content is explained by the two variables?

b. Find the equation of the plane that best fits the data.

c. Predict the amount of tar in a cigarette that contains 1.3 mg of nicotine and 14 mg of carbon monoxide.